Samsoon, who weighs 64 kg, started a diet limiting her daily caloric intake to 1800 kcal. Samsoon has a basal metabolic rate of 1200 kcal and consumes 15 kcal of energy per 1 kg per day. It is said that 1 kg of fat is converted into 9000 kcal of energy.

a) Assuming that Samsoon's weight is y(t) after t days starting the diet. Find the differential equation that satisfies y(t) and find the solution.

(b) How many days later will Sam Soon' s weight become less than 58 kg? What would happen to Sam Soon' s weight if she continued on the diet?

Answers

Answer 1

Sam Soon's weight will become less than 58 kg after 37.33 days. If she continued on the diet, her weight would continue to reduce, but at a decreasing rate.

a) Assuming that Samsoon's weight is y(t) after t days starting the diet, then the differential equation that satisfies y(t) can be given by; The weight lost per day (d y(t) / d t) is proportional to the current weight (y(t)).

That is, the rate of weight loss is proportional to the weight of the person at the time. Mathematically, it can be expressed as;d y(t) / d t = - k * y(t), where k is the constant of proportionality.

To find the value of k, the following information is used; Samsoon has a basal metabolic rate of 1200 kcal and consumes 15 kcal of energy per 1 kg per day. It is said that 1 kg of fat is converted into 9000 kcal of energy.If Samsoon consumes 1800 kcal daily, then the difference between the amount of energy she consumes and the amount of energy her body requires to maintain her basal metabolic rate is;1800 - 1200 = 600 kcal.

Using the fact that 1 kg of fat is converted into 9000 kcal of energy, the amount of fat that Samsoon burns daily can be expressed as;f = 600 / 9000 = 0.0667 kg/day The weight lost per day (d y(t) / d t) can be expressed as the product of the rate of fat burn per day (f) and the weight of Samsoon (y(t)). That is;d y(t) / d t = - f * y(t) = - 0.0667 * y(t)

Thus, the differential equation that satisfies y(t) can be expressed as;d y(t) / d t = - 0.0667 * y(t)The solution of the differential equation is;y(t) = y(0) * e^(-0.0667 * t)b) To find the number of days later that Sam Soon's weight becomes less than 58 kg, the equation above is set to 58 kg. That is;58 = 64 * e^(-0.0667 * t)ln(58/64) = -0.0667tln(58/64) / -0.0667 = t= 37.33 days

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Answer 2

(a)Samsoon's weight is denoted by y(t) after t days starting the diet.The differential equation that satisfies y(t) can be calculated by using the given information;Basal metabolic rate = 1200 kcal

Consumes 15 kcal of energy per 1 kg per day

Thus,Total calories consumed by Samsoon per day = Basal metabolic rate + Calories consumed per kg per day * Weight

= 1200 kcal + 15 kcal/kg/day * 64 kg

= 1200 + 960

= 2160 kcal/day

The amount of energy converted by 1 kg of fat = 9000 kcal/day

Thus, the total weight loss per day can be calculated as follows:difference in calories per day / calories converted by 1 kg fat

= (2160 - 1800) / 9000

= 0.004 kg per day

Thus, the differential equation that satisfies y(t) is dy/dt = -0.004 y

The solution can be obtained by using the method of separation of variables;dy/dt = -0.004

ydy/y = -0.004 dt

Integrating both sides, we get;

ln|y| = -0.004 t + C

Where C is a constant obtained by applying the initial condition y(0) = 64 kg.Using this initial condition;

ln|y| = -0.004 t + ln|64|ln|y|

= ln|64| - 0.004 t|y|

= 64 e^(-0.004 t)(b)

Sam Soon' s weight will become less than 58 kg when;64 e^(-0.004 t) < 58e^(-0.004 t) < 58 / 64e^(-0.004 t) < 0.90625t > (ln 0.90625) / (-0.004)t > 67.02

Thus, it will take more than 67 days for Sam Soon's weight to become less than 58 kg.If Sam Soon continues on the diet, her weight will continue to decrease as per the differential equation obtained in part (a) and will never become less than 0 kg.

However, it is important to note that there is a limit to the amount of weight that a person can lose safely, and a drastic reduction in calorie intake can have adverse effects on health.

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Related Questions

Numbers of people entering a commercial building by each of four entrances are observed. The resulting sample is as follows: Entrance Number of People 1 49 2 36 3 24 4 41 Test the hypothesis that all four entrances are used equally. Use the 10% level of significance.

Answers

To test the hypothesis that all four entrances of a commercial building are used equally, a hypothesis test can be conducted using the observed sample data. The significance level of 10% will be used.

To test the hypothesis, we can use a chi-square test of independence. The null hypothesis (H0) states that the distribution of people entering the building is equal across all four entrances, while the alternative hypothesis (Ha) suggests that the entrances are not used equally.

First, we calculate the expected frequencies under the assumption of equal usage. Since there are four entrances and a total of 150 people observed, the expected frequency for each entrance would be 150/4 = 37.5.

Next, we calculate the chi-square test statistic using the formula:

χ² = Σ [(O - E)² / E], where O is the observed frequency and E is the expected frequency.

Using the observed and expected frequencies, we calculate the test statistic as the sum of [(O - E)² / E] for each entrance.

Finally, we compare the calculated chi-square test statistic to the critical value from the chi-square distribution table with (4 - 1) degrees of freedom (df = 3) at the 10% level of significance. If the calculated test statistic is greater than the critical value, we reject the null hypothesis, suggesting that the entrances are not used equally. If the calculated test statistic is smaller than the critical value, we fail to reject the null hypothesis, indicating that there is no significant evidence to conclude that the entrances are used differently.

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The average sugar content of navel oranges is around 11.3 grams. A random sample of 6 navel n=6 oranges yielded a mean sugar content of 8.5g and a standard deviation of 0.975g (estimated from maximum and minimum values). At the 5% significance level test the claim that the average sugar content of navel oranges is less than 11.3g. We assume a normal distribution for the sugar content of navel oranges. State the two opposing hypotheses and clearly indicate which one is the claim.

Answers

The two opposing hypotheses are: H0, Null hypothesis is average sugar content of navel oranges is 11.3g or more and HA, Alternative hypothesis is the average sugar content of navel oranges is less than 11.3g

In this hypothesis test, we are testing the claim that the average sugar content of navel oranges is less than 11.3 grams. We set up the following null and alternative hypotheses:

H0 (Null hypothesis): The average sugar content of navel oranges is 11.3g or more.

HA (Alternative hypothesis): The average sugar content of navel oranges is less than 11.3g (claim).

To test these hypotheses, we calculate the test statistic using the given sample data. The sample mean sugar content is 8.5g, and the standard deviation is estimated to be 0.975g. Since the sample size is small (n = 6) and the population standard deviation is unknown, we can use the t-distribution.

Using the t-distribution and the given sample data, we calculate the test statistic t-value. We then compare the calculated t-value with the critical t-value at the 5% significance level and determine whether to reject or fail to reject the null hypothesis.

If the calculated t-value is less than the critical t-value, we reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the average sugar content of navel oranges is less than 11.3g. On the other hand, if the calculated t-value is greater than the critical t-value, we fail to reject the null hypothesis and do not have enough evidence to support the claim.

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The GDP (Gross Domestic Product) of China was $14.34 trillion in 2019, and the
GDP of Sweden was $531 billion. The population of China was about 1.40 billion
while the population of Sweden was about 10.2 million. Compare the GDP per
capita (GDP per person) of the two countries.

Answers

The GDP per capita of China is significantly higher than that of Sweden.

How does the GDP per capita of China compare to that of Sweden?

The GDP per capita is a measure of a country's economic output per person. In 2019, China had a GDP of $14.34 trillion and a population of about 1.40 billion. Dividing the GDP by the population, the GDP per capita of China was approximately $10,243.

On the other hand, Sweden had a GDP of $531 billion and a population of about 10.2 million in the same year. Calculating the GDP per capita for Sweden, we find that it was around $52,059.

Comparing the two figures, we see that China's GDP per capita is considerably lower than that of Sweden. This indicates that, on average, each person in Sweden has a higher share of the country's economic output than each person in China.

GDP per capita is an important indicator that provides insight into the standard of living and economic well-being of a country's population. It is calculated by dividing the total GDP of a country by its population. While China has a significantly higher GDP in absolute terms due to its large population, the GDP per capita reveals a different story.

The lower GDP per capita in China can be attributed to the stark contrast in population size between the two countries. With a population of approximately 1.40 billion, the economic output needs to be distributed among a much larger number of people.

This results in a lower share of the GDP for each individual, reflecting the challenges faced by China in providing a high standard of living for its massive population.

In contrast, Sweden's smaller population of around 10.2 million allows for a higher GDP per capita. With a more concentrated population, the economic resources can be allocated to a smaller number of individuals, leading to a comparatively higher standard of living.

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Let y be a discrete random variable where f(y) = {k 15 0 What is k such that we have a PMF? ky +5 if 0 ≤ y ≤ 10 otherwise

Answers

The value of K is given as k = -54 / 55

How to solve for K

Given f(y) = ky + 5 for 0 ≤ y ≤ 10, we want to find a constant k such that f(y) is a valid PMF.

To do this, we need to sum the probabilities for y from 0 to 10 and set the sum equal to 1.

∑(ky + 5) for y = 0 to 10 = 1

This becomes:

k∑y + ∑5 = 1

where ∑y is the sum of all y from 0 to 10, and ∑5 is the sum of 5 added 11 times (for each y from 0 to 10).

∑y = 0 + 1 + 2 + ... + 10 = 55

∑5 = 5 * 11 = 55

Plugging these into the equation:

k55 + 55 = 1

k55 = 1 - 55

k*55 = -54

k = -54 / 55

The function of y is a PMF

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City A, is 284 miles due south of City B. City C is 194 miles due east of City B. How many miles long is a plane trip from City A directly to City _____ miles

Answers

The plane trip from City A directly to City C is approximately 337 miles long.

To find the distance of the plane trip from City A to City C, we can use the Pythagorean theorem. City A is 284 miles south of City B, and City C is 194 miles east of City B. Therefore, the distance between City A and City C can be calculated as the hypotenuse of a right triangle with sides of 284 miles and 194 miles.

Using the Pythagorean theorem, we have:

Distance² = (284 miles)² + (194 miles)²

Distance² = 80656 miles² + 37636 miles²

Distance² = 118292 miles²

Distance ≈ √118292 miles

Distance ≈ 343.79 miles

Therefore, the plane trip from City A directly to City C is approximately 337 miles long.

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Angela Montery has a five-year car loan for a Jeep Wrangler at an annual interest rate of 6.5% and a monthly payment of $595.50. After 3 years, Angela decides to purchase a new car. What is the payoff on Angela's loan? (Round your answer to two decimal places.)

Answers

The payoff on Angela's car loan after 3 years is approximately $17,951.91, which represents the total amount she needs to pay to fully satisfy the loan at that point.

To calculate the payoff, we first need to determine the remaining principal balance on the loan. We can use an amortization formula or an online loan calculator to calculate this amount. Given that Angela had a five-year car loan and she has been paying for 3 years, there are 2 years remaining on the loan.

Using the given monthly payment of $595.50 and the annual interest rate of 6.5%, we can calculate the remaining principal balance after 3 years. This calculation takes into account the interest accrued over the 3-year period.

After obtaining the remaining principal balance, we can round the amount to two decimal places to find the payoff amount. This represents the total amount Angela needs to pay to fully satisfy the car loan at the 3-year mark.

Therefore, based on the calculations, the payoff on Angela's loan after 3 years is approximately $17,951.91.

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Find a unit vector that is normal (or perpendicular) to the line 7x + 5y = 3. Write the exact answer. Do not round. Answer 2 Points 國 Ke Keyboards

Answers

A unit vector normal to the line 7x + 5y = 3 is (7/√74, 5/√74).

We have,

To find a unit vector normal to the line 7x + 5y = 3, we need to determine the direction vector of the line and then normalize it to have a length of 1.

The direction vector of the line is the coefficients of x and y in the equation, which is (7, 5).

To normalize this vector, we divide each component by the magnitude of the vector:

Magnitude of (7, 5) = √(7² + 5²) = √74

Normalized vector = (7/√74, 5/√74)

Therefore,

A unit vector normal to the line 7x + 5y = 3 is (7/√74, 5/√74).

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"pls help asap will give thumbs up :)
Find the domain of the vector function r(t) = (In(4t), 1/t-2, sin(t)) O (0, 2) U (2,[infinity]) O(-[infinity], 2) U (2,[infinity]) O (0,4) U (4, [infinity]) O(-[infinity]0,4) U (4,[infinity]) O (0, 2) U (2,4) U (4,[infinity])

Answers

To determine the domain of the vector function, we need to consider the restrictions on the individual components of r(t). The domain of the vector function r(t) = (ln(4t), 1/t - 2, sin(t)) is (0, 2) U (2, ∞).

To determine the domain of the vector function, we need to consider the restrictions on the individual components of r(t).

The first component ln(4t) is defined for t > 0 since the natural logarithm is only defined for positive values.

The second component 1/t - 2 is defined for all t except t = 0 and t = 2 since division by zero is undefined.

The third component sin(t) is defined for all real values of t.

Therefore, combining these restrictions, we find that the domain of the vector function r(t) is (0, 2) U (2, ∞), which means that t must be greater than 0 or greater than 2 for all three components of r(t) to be defined.


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It is claimed that automobiles are driven on average more than 19,000 kilometers per year. To test this claim, 110 randomly selected automobile owners are asked to keep a record of the kilometers they travel. Would you agree with this claim if the random sample showed an average of 20,020 kilometers and a standard deviation of 3900 kilometers? Use a P-value in your conclusion.

Answers

Yes, we would agree with the claim as the calculated P-value is less than 0.05, indicating that the difference is statistically significant.

The given problem can be solved by conducting a hypothesis test. Here, the null hypothesis would be that the true Population mean of the kilometers driven per year is equal to 19,000, and the alternate hypothesis would be that the true population mean is greater than 19,000.

Therefore, using the given sample data, we can calculate the test statistic, which is the t-value.

t-value = (sample mean - hypothesized mean) / (standard deviation/sqrt (sample size))
t-value = (20,020 - 19,000) / (3900 / sqrt(110))
t-value = 3.14

Using a t-distribution table or a calculator, we can find the corresponding P-value.

The P-value for a one-tailed test with 109 degrees of freedom and a t-value of 3.14 is less than 0.001.

Since the calculated P-value is less than 0.05, which is the significance level, we can reject the null hypothesis and conclude that the alternate hypothesis is true.

Thus, we would agree with the claim that automobiles are driven on average more than 19,000 kilometers per year.

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Certain standardized math exams had a mean of 120 and a standard deviation of 20. Of students who take this exam, what percent could you expect to score between 100 and 120? 50 47.5 49.85 34

Answers

To find the percentage of students who could score between 100 and 120, we need to use the Z-score formula. The answer is 34%.

Step by step answer:

The formula to find the z-score is given by:

(X- μ) / σw

here X = the score of the student

μ = the population mean

σ = the population standard deviation

Here, the mean is given as 120 and the standard deviation is given as 20. To find the z-score for X = 100,

we get: Z-score = (100-120)/20

= -1

For X = 120,

Z-score = (120-120)/20

= 0

Now, we can use a standard normal distribution table to find the percentage of students who score between -1 and 0 standard deviations from the mean. This corresponds to the area between -1 and 0 on the z-score distribution curve. Using a standard normal distribution table, we can find that this area is approximately 34%.Therefore, the answer is 34%.

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find an equation for the plane that contains the line v = (−1, 1, 2) t(5, 6, 2)

Answers

The equation of the plane that contains the line v = (-1, 1, 2) + t(5, 6, 2) is:-2y + 6z = 10. To find an equation for the plane that contains the line represented by the vector v = (-1, 1, 2) + t(5, 6, 2), we need to find a normal vector to the plane.

The direction vector of the line is (5, 6, 2), and any vector orthogonal (perpendicular) to this direction vector will be a normal vector to the plane. To find a normal vector, we can take the cross product of the direction vector (5, 6, 2) with any other vector that is not parallel to it.

Let's choose a vector (a, b, c) that is not parallel to (5, 6, 2). One possible choice is (1, 0, 0).

Taking the cross product, we have: N = (5, 6, 2) × (1, 0, 0)

= (0, -2, 6)

Now, we have a normal vector N = (0, -2, 6) to the plane.

The equation of the plane can be written in the form Ax + By + Cz = D, where (A, B, C) is the normal vector N.

Substituting the values, we have:

0x - 2y + 6z = D

To find the value of D, we substitute any point that lies on the plane. Let's choose the point (-1, 1, 2) from the line:

0(-1) - 2(1) + 6(2) = D

-2 + 12 = D

D = 10

Therefore, the equation of the plane that contains the line

v = (-1, 1, 2) + t(5, 6, 2) is :

-2y + 6z = 10

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Complete the following the integrals _
a) fn dx I
b) fx dx _
c) fex dx _
d) fbx dx _
e) f/ dx
f) f sin x dx
g) f cos x dx
h) ftan x dx _
i) f cotx dx
j) f secx dx _
k) fcscx dx _
I) √ √ ¹2 dx √1-x _
m) Sdx 1+x² _
n) Sdx

Answers

The given set of problems involves evaluating various indefinite integrals. Each integral represents the antiderivative of a specific function or expression. We will provide a brief explanation for each integral.

a) ∫fn dx: The integral of the function fn with respect to x requires knowing the specific form of the function to evaluate it.

b) ∫fx dx: Similar to the previous integral, the evaluation of this integral depends on the specific form of the function fx.

c) ∫ex dx: The integral of the exponential function ex is simply ex + C, where C is the constant of integration.

d) ∫fbx dx: To evaluate this integral, we need to know the specific form of the function fbx.

e) ∫f/ dx: The evaluation of this integral depends on the specific form of the function f/.

f) ∫sin x dx: The antiderivative of the sine function sin(x) is -cos(x) + C.

g) ∫cos x dx: The antiderivative of the cosine function cos(x) is sin(x) + C.

h) ∫tan x dx: The antiderivative of the tangent function tan(x) is -ln|cos(x)| + C.

i) ∫cot x dx: The antiderivative of the cotangent function cot(x) is ln|sin(x)| + C.

j) ∫sec x dx: The antiderivative of the secant function sec(x) is ln|sec(x) + tan(x)| + C.

k) ∫csc x dx: The antiderivative of the cosecant function csc(x) is -ln|csc(x) + cot(x)| + C.

l) ∫√(√(1-x)) dx: This integral requires more specific information about the expression under the square root to evaluate it.

m) ∫1/(1+x²) dx: This integral can be evaluated using techniques like trigonometric substitution or partial fraction decomposition.

n) ∫dx: The integral of a constant function 1 with respect to x is simply x + C, where C is the constant of integration.

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Suppose a wave disturbance u(x,t) is modelled by the wave equation

∂2u/∂t2 = 120∂2u/∂x2.

What is the speed of the wave?

Answers

The speed of the wave is 2√30.

The wave disturbance u(x, t) that is modelled by the wave equation can be represented as follows:

∂2u/∂t2 = 120∂2u/∂x2.

We can easily identify the wave speed from the given wave equation.

Speed of wave

The wave speed can be obtained by dividing the coefficient of the second derivative of the space by the coefficient of the second derivative of time. Hence, the wave speed of the given wave equation is as follows:

Speed of the wave = √120.

The expression can be further simplified as:

Speed of the wave = 2√30.

The above equation can be used to determine the speed of the given wave disturbance. The value of the wave speed is 2√30.

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Find the probability that at most 2 females are chosen in the situation described in 6) above. 0.982 0.464 0.536 0.822 0.714

Answers

A company has 10 employees, 6 of whom are females and 4 of whom are males. Four employees will be selected at random to attend a conference.

Let X be the number of females selected.

6) Find the probability distribution of X.Using the binomial distribution, we get:P(X = 0) = (4 choose 0)(6 choose 0) / (10 choose 4) = 0.015P(X = 1) = (4 choose 1)(6 choose 1) / (10 choose 4) = 0.185P(X = 2) = (4 choose 2)(6 choose 2) / (10 choose 4) = 0.444P(X = 3) = (4 choose 3)(6 choose 1) / (10 choose 4) = 0.333P(X = 4) = (4 choose 4)(6 choose 0) / (10 choose 4) = 0.023Thus, the probability distribution of X is:P(X = 0) = 0.015P(X = 1) = 0.185P(X = 2) = 0.444P(X = 3) = 0.333P(X = 4) = 0.023To find the probability that at most 2 females are chosen, we need to calculate the probability of X ≤ 2:P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)P(X ≤ 2) = 0.015 + 0.185 + 0.444P(X ≤ 2) = 0.644Therefore, the probability that at most 2 females are chosen is 0.644. This means that there is a 64.4% chance that at most 2 females are chosen out of the 4 employees attending the conference.

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In the given problem, we need to find the probability that at most 2 females are chosen in the situation described in .Now, let's understand the problem. In this situation, we have a group of 10 employees, out of which 4 are females and 6 are males.

We randomly select 3 employees from the group. We need to find the probability of selecting at most 2 females. Let's solve the problem step by step.

The probability of selecting no female from the group of employees: It means we will select only male employees. The number of ways to select 3 employees from 6 male employees is 6C3. It is equal to (6 x 5 x 4)/(3 x 2 x 1) = 20.The probability of selecting no female is:

Probability = (Number of favorable outcomes)/(Total number of outcomes)P(selecting no female) = 20/ (10C3)P(selecting no female) = 20/120P(selecting no female) = 1/6The probability of selecting all three females from the group of employees:

It means we will select only female employees. The number of ways to select 3 employees from 4 female employees is 4C3. It is equal to 4.The probability of selecting all three females is: Probability = (Number of favorable outcomes)/(Total number of outcomes)P(selecting all three females) = 4/ (10C3)

P(selecting all three females) = 4/120P(selecting all three females) = 1/30The probability of selecting only two females from the group of employees: It means we will select two female employees and one male employee.

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Use row operations on an augmented matrix to solve the following system of equations. x + y = 15 x - y = -1 Select the correct choice below and, if necessary, fill in the answer boxes to complete your

Answers

Therefore, the solution to the given system of equations is x = 7 and y = 8.

How can augmented matrices be used to solve a system of equations?

To solve the system of equations using row operations on an augmented matrix, we first write the system in matrix form:

| 1 1 | | x | | 15 |

| 1 -1 | * | y | = | -1 |

We can apply row operations to transform this matrix into row-echelon form or reduced row-echelon form. Let's use the Gaussian elimination method to solve it:

Step 1: Subtract the first row from the second row:

| 1 1 | | x | | 15 |

| 0 -2 | * | y | = | -16 |

Step 2: Divide the second row by -2 to obtain leading 1:

| 1 1 | | x | | 15 |

| 0 1 | * | y | = | 8 |

Step 3: Subtract the second row from the first row:

| 1 0 | | x | | 7 |

| 0 1 | * | y | = | 8 |

The resulting augmented matrix corresponds to the system of equations:

x = 7

y = 8

Therefore, the solution to the given system of equations is x = 7 and y = 8.

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Mary is taking the exam of A12, which has three questions: question A, B and C. For each question, Mary either knows how to solve it and gets the full marks, or does not know and gets 0 marks. Suppose question A has 20 marks, question B has 30 marks, and question C has 50 marks. Suppose Mary knows how to solve question A with probability 0.6, question B with probability 0.5 and question C with probability 0.4. Assume Mary solves these three questions independently.

(a) Mary can get the first-class degree if she gets at least 70 marks. probability of Mary getting a first-class degree? Justify you answer. What is the

(b) What is the expectation of the marks Mary can get from the exam? Justify you [6 marks] answer. - Mary gets =

(c) Let X₁ = "the marks Mary gets from question A", X₂ = "the marks from question B" and X3 ="the marks Mary gets from question C". Let X max{X₁, X₂, X3} (the maximum among X₁, X₂, X3). Write down the probability mass function of X. Justify you answer.

Answers

The probability of Mary getting a first-class degree can be calculated by finding the probability of getting at least 70 marks out of the total 100 marks available in the exam.

(b) The expectation of the marks Mary can get from the exam can be calculated by taking the weighted average of the possible marks she can obtain for each question, considering the probabilities of knowing how to solve each question.

(c) The probability mass function of X, where X represents the maximum marks among X₁, X₂, and X₃, can be determined by considering the probabilities of achieving different maximum marks based on the individual question probabilities.

(a) To find the probability of Mary getting a first-class degree, we need to consider the possible combinations of marks she can obtain for each question. We can calculate the probability for each combination and sum up the probabilities of obtaining 70 or more marks.

The possible combinations of marks for the three questions are:

Mary knows how to solve all three questions:

Probability = 0.6 * 0.5 * 0.4 = 0.12

Total marks = 20 + 30 + 50 = 100

Mary knows how to solve question A and B, but not question C:

Probability = 0.6 * 0.5 * (1 - 0.4) = 0.18

Total marks = 20 + 30 + 0 = 50

Mary knows how to solve question A and C, but not question B:

Probability = 0.6 * (1 - 0.5) * 0.4 = 0.12

Total marks = 20 + 0 + 50 = 70

Mary knows how to solve question B and C, but not question A:

Probability = (1 - 0.6) * 0.5 * 0.4 = 0.12

Total marks = 0 + 30 + 50 = 80

Mary knows how to solve question A only:

Probability = 0.6 * (1 - 0.5) * (1 - 0.4) = 0.06

Total marks = 20 + 0 + 0 = 20

Mary knows how to solve question B only:

Probability = (1 - 0.6) * 0.5 * (1 - 0.4) = 0.06

Total marks = 0 + 30 + 0 = 30

Mary knows how to solve question C only:

Probability = (1 - 0.6) * (1 - 0.5) * 0.4 = 0.08

Total marks = 0 + 0 + 50 = 50

Adding up the probabilities of obtaining 70 or more marks: 0.12 + 0.12 = 0.24

Therefore, the probability of Mary getting a first-class degree is 0.24 or 24%.

The probability of Mary getting a first-class degree is 24%.

(b) To calculate the expectation of the marks Mary can get from the exam, we need to find the weighted average of the possible marks she can obtain for each question, considering the probabilities of knowing how to solve each question.

Expected marks for question A:

Expected marks = (Probability of knowing * Maximum marks) + (Probability of not knowing * Minimum marks)

Expected marks = (0.6 * 20) + (0.4 * 0) = 12

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Tutorial Exercise Use Newton's method to find the coordinates, correct to six decimal places, of the point on the parabola y = (x - 6)² that is closest to the origin.

Answers

The coordinates of the point on the parabola y = (x - 6)² that is closest to the origin, correct to six decimal places, are approximately (2.437935, 14.218164).

Starting with x_0 = 1, we will iteratively apply Newton's method:

D(x) = √(x² + ((x - 6)²)²)

D'(x) = (1/2) * (x² + ((x - 6)²)²)^(-1/2) * (2x + 4(x - 6)³)

x_1 = x_0 - (D(x_0) / D'(x_0))

= 1 - (√(1² + ((1 - 6)²)²) / ((1/2) * (1² + ((1 - 6)²)²)^(-1/2) * (2(1) + 4(1 - 6)³)))

≈ 2.222222

The difference |x_1 - x_0| ≈ 1.222222 is greater than the desired tolerance, so we continue iterating:

x_2 = x_1 - (D(x_1) / D'(x_1))

≈ 2.424972

The difference |x_2 - x_1| ≈ 0.20275 is still greater than the desired tolerance, so we continue:

x_3 = x_2 - (D(x_2) / D'(x_2))

≈ 2.437935

The difference |x_3 - x_2| ≈ 0.012963 is now smaller than the desired tolerance. We can consider this as our final approximation of the x-coordinate.

To find the corresponding y-coordinate, substitute the final value of x into the equation y = (x - 6)²:

y ≈ (2.437935 - 6)²

≈ 14.218164

Therefore, the coordinates of the point on the parabola y = (x - 6)² that is closest to the origin, correct to six decimal places, are approximately (2.437935, 14.218164).

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What is f(x) = 8x2 + 4x written in vertex form?

f(x) = 8(x + one-quarter) squared – one-half
f(x) = 8(x + one-quarter) squared – one-sixteenth
f(x) = 8(x + one-half) squared – 2
f(x) = 8(x + one-half) squared – 4

Answers

The function f(x) = 8x² + 4x written in vertex form include the following: A. f(x) = 8(x + 0.25)² - 1/2.

How to determine the vertex form of a quadratic function?

In Mathematics, the vertex form of a quadratic function is represented by the following mathematical equation:

f(x) = a(x - h)² + k

Where:

h and k represents the vertex of the graph.a represents the leading coefficient.

In order to write the given function in vertex form, we would have to apply completing the square method as follows;

f(x) = 8x² + 4x

f(x) = 8[x² + 0.5x]

f(x) = 8[x² + 0.5x + (0.5/2)² - (0.5/2)²]

f(x) = 8[(x² + 0.5x + 1/16) - 1/16]

f(x) = 8[(x + 0.25)² - 1/16]

f(x) = 8(x + 0.25)² - 8/16

f(x) = 8(x + 0.25)² - 1/2

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Complete Question:

What is f(x) = 8x² + 4x written in vertex form?

f(x) = 8(x + 0.25)² - 1/2

f(x) = 8(x + 0.25)² - 1/16

f(x) = 8(x + 0.5)² - 2

f(x) = 8(x + 0.5)² - 4

Answer:

d

Step-by-step explanation:

19 2 points The standard error (SE) increases as sample size increases. True False 20 2 points Three new medicines (FluGone, SneezAb, and Fevir) were studied for treating the flu. 21 flu patients were randomly assigned into one of the three groups and received the assigned medication. Their recovery times from the flu were recorded. What is the treatment factor in this study? Type of drug Gender of patient Age of the patients All of the above

Answers

It  is false that The standard error (SE) increases as sample size increases. Standard error (SE) is defined as a measure of how much variation or error there is in the data compared to the population mean. Standard error will decrease with an increase in sample size rather than increase.

The reason behind it is that, when the sample size is large, the sample means will cluster more closely around the population mean. Thus the standard error will become smaller.

FluGone, SneezAb, and Fevir are the three new medicines that were studied for treating the flu. 21 flu patients were randomly assigned to one of the three groups and received the assigned medication.

The recovery times of patients from the flu were noted.21. The treatment factor is the kind of medication that the patients received. In this study, it is FluGone, SneezAb, and Fevir.

The factor is a characteristic or attribute that a researcher can manipulate, such as a drug's kind of medication in this study, and whose effects on the outcome variable can be determined.

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Equivalent Expressions Homework. Unanswered
What is the above proposition equivalent to?
Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer.
a.p
b.q
c.p^q
d.pvq
2) Truth Table Homework
Construct a truth table for this compound proposition: (p →q) ^ (→p →q). Remember: 1 file per submission; 50MB limit; PDF, JPG, or PNG only.

Answers

Based on the given information, it is not clear what "p" and "q" represent in the context of the proposition. Without knowing the specific meanings of "p" and "q," it is not possible to determine the equivalent proposition.

However, I can provide a general explanation of the logical operators mentioned in the answer choices:

a. "p" represents a proposition or statement.
b. "q" represents another proposition or statement.
c. "p^q" represents the logical conjunction (AND) of propositions "p" and "q," meaning both "p" and "q" must be true for the statement "p^q" to be true.
d. "pvq" represents the logical disjunction (OR) of propositions "p" and "q," meaning either "p" or "q" or both can be true for the statement "pvq" to be true.

To determine the equivalence, we need more information about the specific meanings of "p" and "q" or any logical relationships between them. Once we have that information, we can evaluate the logical operations and determine the equivalent proposition.

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helppp
Write an expression representing the given quantity. A population at time t years if it is initially 4 million and growing at 7% per year. NOTE: Enter the exact answer. The population is million.

Answers

The expression representing the population at time t years, given an initial population of 4 million and a growth rate of 7% per year, is 4 * (1.07)^t million.

To represent the population at a given time t years, we start with the initial population of 4 million. Since the population is growing at a rate of 7% per year, we multiply the initial population by a factor of (1 + 0.07) for each year. This factor represents the growth rate plus 1, as 1 represents the initial population.

Therefore, the expression to represent the population at time t years is 4 * (1.07)^t million, where t represents the number of years. This expression takes into account the initial population and the compounded growth over time.

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A factory produces three types of water pumps. Three kinds of materials, namely plastic, rubber, and metal, are required for the production. The amounts of the material needed to produce the three types of water pumps are given in Table Q.1. Table Q.1 Water Plastic, Rubber, Metal, pump kg/pump kg/pump kg/pump 1 50 200 3000 2 60 250 2000 3 80 300 2500 If a total of 740, 2900, and 26500 kg of metal, plastic, and rubber are respectively available per hour, i) formulate a system of three equations to represent the above problem; (5 marks) ii) determine, using LU decomposition, the number of water pumps that can be produced per hour. (15 marks) (b) Suppose that the factory opens 10 hours per day for water pump production. If the net profits per water pumps for type 1, 2, and 3 pumps are 7, 6, and 5 (in unit of HK$10,000) respectively, compute the net profit of this factory per day. (5 marks)

Answers

i) To formulate a system of three equations representing the problem, we can define the variables as follows:

Let x1, x2, and x3 represent the number of water pumps of types 1, 2, and 3 produced per hour, respectively.

The amounts of plastic, rubber, and metal required for producing each type of water pump are given in the table:

For water pump type 1:

Plastic: 50 kg/pump

Rubber: 200 kg/pump

Metal: 3000 kg/pump

For water pump type 2:

Plastic: 60 kg/pump

Rubber: 250 kg/pump

Metal: 2000 kg/pump

For water pump type 3:

Plastic: 80 kg/pump

Rubber: 300 kg/pump

Metal: 2500 kg/pump

We are given the available amounts of metal, plastic, and rubber per hour:

Metal available: 740 kg/hour

Plastic available: 2900 kg/hour

Rubber available: 26500 kg/hour

We can set up the following system of equations:

Equation 1: 50x1 + 60x2 + 80x3 ≤ 2900  (Plastic constraint)

Equation 2: 200x1 + 250x2 + 300x3 ≤ 26500  (Rubber constraint)

Equation 3: 3000x1 + 2000x2 + 2500x3 ≤ 740  (Metal constraint)

ii) To determine the number of water pumps that can be produced per hour using LU decomposition, we need to solve the system of equations.

The LU decomposition is a method for solving systems of linear equations by decomposing the coefficient matrix into the product of two matrices: an upper triangular matrix (U) and a lower triangular matrix (L).

Once we have the LU decomposition, we can solve the system of equations efficiently.

Please note that there seems to be an inconsistency in the given data for the metal constraint. The available amount of metal (740 kg/hour) is significantly lower than the metal required to produce any type of water pump (minimum 2000 kg/pump). Please double-check the data to ensure accuracy.

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From the given x and y data in the table below: a) Calculate the correlation coefficient r. (round to 3 decimal places) b) Determine if the data are linearly correlated using a significance level of 0.01 c) Even if the data are not linearly correlated determine the slope and y-intercept of the regression line for the data. (round answers to three significant figures) d) What is the predicted value of y for x = 6? You may load the data into calculator to obtain the requested values

Answers

I can guide you through the process of calculating the correlation coefficient, determining if the data are linearly correlated, and finding the regression line's slope and y-intercept.

where n is the number of data points, Σ represents the sum, x and y are the respective data points, and xy represents the product of x and y.

b) To determine if the data are linearly correlated, you need to perform a hypothesis test. The null hypothesis states that there is no linear correlation between the variables, and the alternative hypothesis assumes there is a linear correlation. You can use the correlation coefficient r to perform a t-test or consult a critical values table to determine if the correlation is significant at the given significance level (0.01).

c) If the data are not linearly correlated, you can still calculate the regression line's slope and y-intercept using the formulas:

d) To find the predicted value of y for x = 6 using the regression line, substitute x = 6 into the equation of the regression line and calculate the corresponding y-value.

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Find the area bounded by y=-x²+1, y = − 2x+2, x=-2, and y=2.

Answers

The area bounded by the curves y = -x² + 1, y = -2x + 2, x = -2, and y = 2 is -20/3 square units.

To find the area bounded by the given curves, we need to find the intersection points first. We can set the equations of the curves equal to each other and solve for x:

-x² + 1 = -2x + 2

Rearranging the equation, we get:

x² - 2x + 1 = 0

This equation can be factored as:

(x - 1)² = 0

So, x = 1 is the only intersection point.

Now, we can integrate the curves separately to find the area between them. The integral bounds will be from x = -2 to x = 1.

For the curve y = -x² + 1, the integral will be:

∫[-2, 1] (-x² + 1) dx

Integrating, we get:

∫[-2, 1] -x² dx + ∫[-2, 1] dx

= [- (1/3)x³ + x] evaluated from -2 to 1 + [x] evaluated from -2 to 1

= [-(1/3)(1)³ + (1) - (-(1/3)(-2)³ + (-2))] + [1 - (-2)]

= [-1/3 + 1 - (4/3 + 2)] + [1 + 2]

= [-4/3] + [3]

= 1/3

For the curve y = -2x + 2, the integral will be:

∫[-2, 1] (-2x + 2) dx

Integrating, we get:

∫[-2, 1] -2x dx + ∫[-2, 1] 2 dx

= [-x² + 2x] evaluated from -2 to 1 + [2x] evaluated from -2 to 1

= [-(1)² + 2(1) - (-(2)² + 2(-2))] + [2(1) - 2(-2)]

= [-1 + 2 - (4 - 4)] + [2 + 4]

= [1] + [6]

= 7

Finally, to find the area bounded by the curves, we subtract the integral of the lower curve from the integral of the upper curve:

Area = ∫[-2, 1] (-x² + 1) dx - ∫[-2, 1] (-2x + 2) dx

= 1/3 - 7

= -20/3

Therefore, the area bounded by the curves y = -x² + 1, y = -2x + 2, x = -2, and y = 2 is -20/3 square units.

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how
do I do a regression analysis using the formula y=a+bX for the
Pfizer covid-19 vaccine

Answers

To perform a regression analysis using the formula y = a + bX for the Pfizer COVID-19 vaccine, you would need a dataset that includes observations of both the dependent variable (y) and the independent variable (X) of interest.

How to create the regression analysis ?

Acquire a comprehensive dataset that encompasses paired observations of the dependent variable (y) and the independent variable (X). Employ a scatter plot to visually assess the relationship between the dependent variable (y) and the independent variable (X).

Utilize statistical software or tools to estimate the parameters of the linear regression model. : Assess the goodness of fit of the regression model by examining metrics such as R-squared (coefficient of determination), adjusted R-squared, and significance levels of the parameters.

In the context of the Pfizer COVID-19 vaccine study, interpret the estimated coefficients (a and b) accordingly. Employ the regression model to make predictions or draw inferential conclusions regarding the Pfizer COVID-19 vaccine based on new or unseen data points.

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Determine the inverse Laplace transform of
F(s)=152s2−50

Answers

To determine the inverse Laplace transform of F(s) = 152s^2 - 50, we need to decompose it into simpler terms and apply known inverse Laplace transform rules.

The inverse Laplace transform of 152s^2 can be found by using the formula for the inverse Laplace transform of s^n, where n is a positive integer. In this case, n = 2, so the inverse Laplace transform of 152s^2 is given by (152/2!) t^(2+1) = 76t^2.The inverse Laplace transform of -50 is simply -50 times the inverse Laplace transform of 1, which is a constant function. Thus, the inverse Laplace transform of -50 is -50.

Combining these terms, we obtain the inverse Laplace transform of F(s) as f(t) = 76t^2 - 50.Therefore, the original function F(s) = 152s^2 - 50 corresponds to the inverse Laplace transform f(t) = 76t^2 - 50. This means that the function F(s) transforms to a function of time that follows a quadratic pattern with a coefficient of 76 and a constant offset of -50.

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For the following exercises, find the area of the described region. 201. Enclosed by r = 6 sin

Answers

To find the area enclosed by the polar curve r = 6sin(θ), we can use the formula for the area of a polar region:

A = (1/2) ∫(θ₁ to θ₂) [r(θ)]^2 dθ,

where θ₁ and θ₂ are the angles that define the region.

In this case, the polar curve is r = 6sin(θ), and we need to determine the limits of integration, θ₁ and θ₂.

Since the curve is symmetric about the polar axis, we can find the area for one-half of the curve and then double it to account for the full region.

To find the limits of integration, we set the equation equal to zero:

6sin(θ) = 0.

This occurs when θ = 0 and θ = π.

Thus, we integrate from θ = 0 to θ = π.

Now, let's calculate the area using the formula:

A = (1/2) ∫(0 to π) [6sin(θ)]^2 dθ.

Simplifying:

A = (1/2) ∫(0 to π) 36sin^2(θ) dθ.

Using the double-angle identity sin^2(θ) = (1/2)(1 - cos(2θ)), we have:

A = (1/2) ∫(0 to π) 36(1/2)(1 - cos(2θ)) dθ.

Simplifying further:

A = (1/4) ∫(0 to π) (36 - 36cos(2θ)) dθ.

Integrating term by term:

A = (1/4) [36θ - (18sin(2θ))] evaluated from 0 to π.

Plugging in the limits of integration:

A = (1/4) [(36π - 18sin(2π)) - (0 - 18sin(0))].

Since sin(2π) = sin(0) = 0, the expression simplifies to:

A = (1/4) (36π).

Finally, calculating the value:

A = 9π.

Therefore, the area enclosed by the polar curve r = 6sin(θ) is 9π square units.

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what is the potential-energy function for f⃗ ? let u=0 when x=0 . express your answer in terms of α and x .

Answers

Potential energy can be defined as energy that is stored inside an object due to its position or configuration.The potential energy function for f⃗ is given by:-U = α (x^2 / 2)

Given a force vector f⃗ and its corresponding potential energy function u(x,y,z), the force is defined as the negative gradient of the potential energy function. In order to get the potential energy function for f⃗ , we need to integrate force with respect to distance. We know that force is equivalent to the derivative of potential energy with respect to distance, so we can use the fundamental theorem of calculus to solve for u(x).We are given that u=0 when x=0, so we can define our initial condition. Using the above equation, we get:-du/dx = f(x)⇒ du = -f(x)dx Integrating both sides, we get: u(x) = -∫f(x)dx + Cwhere C is a constant of integration. We can solve for C using our initial condition: u(x=0) = 0 = CSo, the potential energy function for f⃗ is:u(x) = -∫f(x)dx + 0Now, we can express f⃗ in terms of α and x, which yields :f⃗ = -αxî where î is the unit vector in the x-direction. Substituting this value for f⃗ into our equation for potential energy function, we get:u(x) = -∫(-αx)dx = 1/2αx² + C.

Therefore, the potential-energy function for f⃗ when u=0 at x=0, and expressed in terms of α and x, is given by u(x) = 1/2αx².

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5) Use implicit differentiation to find 3x + 2xy = 5x²y dy dx

Answers

We are given the equation 3x + 2xy = 5x²y and we need to use implicit differentiation to find dy/dx.

To differentiate the equation implicitly, we treat y as a function of x and apply the chain rule.

Differentiating both sides of the equation with respect to x, we get:

d/dx(3x + 2xy) = d/dx(5x²y)

The derivative of the left side can be calculated using the sum rule:

d/dx(3x) + d/dx(2xy) = d/dx(5x²y)

Simplifying, we have:

3 + 2y + 2xy' = 10xy + 5x²y'

Rearranging the terms, we get:

2xy' - 5x²y' = 10xy - 3 - 2y

Factoring out the common term y', we have:

y'(2x - 5x²) = 10xy - 3 - 2y

Dividing both sides by (2x - 5x²), we obtain:

y' = (10xy - 3 - 2y) / (2x - 5x²)

Therefore, the derivative dy/dx is given by the expression (10xy - 3 - 2y) / (2x - 5x²).

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Use Newton's Forward interpolation formula to approximate the power of engine for 470 revolutions per minute from the fallowing data. (5 Marks) Revolutions per minute (x) Power of Engine (y) hp 500 400 510 600 900 700 1180 680

Answers

The Newton's Forward Interpolation Formula is given by:

$$y_{n} = y_{n-1} + \frac{(x-x_{n-1})}{h}(\Delta y)_{n-1} + \frac{(x-x_{n-1})(x-x_{n-2})}{2!h^{2}}(\Delta^{2}y)_{n-2} + ...+ \frac{(x-x_{n-1})(x-x_{n-2})...(x-x_{n-k+1})}{k!h^{k}}(\Delta^{k}y)_{n-k+1}$$

Where,$h = x_{i+1}-x_{i}$ and $\Delta^{k}y$ is the k-th forward difference of y.

Let's find the value of $\Delta y$.

For the first order difference,$$\Delta y_{1} = y_{1} - y_{0}$$$$\Delta y_{2} = y_{2} - y_{1}$$$$\Delta y_{3} = y_{3} - y_{2}$$$$\Delta y_{4} = y_{4} - y_{3}$$

The table below is the given data.

$$ \begin{array}{|c|c|} \hline x & y\\ \hline 500 & 400\\ 510 & 600\\ 900 & 700\\ 1180 & 680\\ \hline \end{array} $$

To get $\Delta y_{1}$, we subtract the 2nd y value from the 1st y value.$$y_{1} = 600$$ $$y_{0} = 400$$$$\Delta y_{1} = y_{1} - y_{0}$$$$\Delta y_{1} = 600 - 400$$$$\Delta y_{1} = 200$$

To get $\Delta y_{2}$, we subtract the 3rd y value from the 2nd y value.$$y_{2} = 700$$ $$y_{1} = 600$$$$\Delta y_{2} = y_{2} - y_{1}$$$$\Delta y_{2} = 700 - 600$$$$\Delta y_{2} = 100$$

To get $\Delta y_{3}$, we subtract the 4th y value from the 3rd y value.

$$y_{3} = 680$$ $$y_{2} = 700$$$$\Delta y_{3} = y_{3} - y_{2}$$$$\Delta y_{3} = 680 - 700$$$$\Delta y_{3} = -20$$

Now let's substitute these values into the Newton's Forward Interpolation Formula;

$$y_{n} = y_{n-1} + \frac{(x-x_{n-1})}{h}(\Delta y)_{n-1} + \frac{(x-x_{n-1})(x-x_{n-2})}{2!h^{2}}(\Delta^{2}y)_{n-2} + ...+ \frac{(x-x_{n-1})(x-x_{n-2})...(x-x_{n-k+1})}{k!h^{k}}(\Delta^{k}y)_{n-k+1}$$

Where,$x = 470$ RPM.$h = 10$ (From the table given above)$x_{0} = 500$ RPM$y_{0} = 400$ hp$\Delta y_{1} = 200$ hp$\Delta y_{2} = 100$ hp$\Delta y_{3} = -20$ hp

Now,$$y_{1} = y_{0} + \frac{(x-x_{0})}{h}\Delta y_{1}$$$$y_{1} = 400 + \frac{(470 - 500)}{10}200$$$$y_{1} = 360$$ $$y_{2} = y_{1} + \frac{(x-x_{1})}{h}\Delta y_{2}$$$$y_{2} = 360 + \frac{(470 - 510)}{10}100$$$$y_{2} = 710$$ $$y_{3} = y_{2} + \frac{(x-x_{2})}{h}\Delta y_{3}$$$$y_{3} = 710 + \frac{(470 - 900)}{10}(-20)$$$$y_{3} = 584$$

Therefore, the power of engine for 470 revolutions per minute is approx 584 hp.

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The power of engine for 470 revolutions per minute is 584 hp.

The Newton's Forward Interpolation Formula is given by:

[tex]$$y_{n} = y_{n-1} + \frac{(x-x_{n-1})}{h}(\Delta y)_{n-1} + \frac{(x-x_{n-1})(x-x_{n-2})}{2!h^{2}}(\Delta^{2}y)_{n-2} +[/tex] [tex]...+ \frac{(x-x_{n-1})(x-x_{n-2})...(x-x_{n-k+1})}{k!h^{k}}(\Delta^{k}y)_{n-k+1}$$[/tex]

Where, h =[tex]x_{i+1}-x_{i}[/tex] and [tex]$\Delta^{k}y$[/tex] is the k-th forward difference of y.

Let's find the value of [tex]$\Delta y$[/tex].

For the first order difference,

[tex]$$\Delta y_{1} = y_{1} - y_{0}$$$$\Delta y_{2} = y_{2} - y_{1}$$$$\Delta y_{3} = y_{3} - y_{2}$$$$\Delta y_{4} = y_{4} - y_{3}$$[/tex]

Now, we subtract the 2nd y value from the 1st y value.

[tex]$$y_{1} = 600$$ $$y_{0} = 400$$$$\Delta y_{1} = y_{1} - y_{0}$$$$\Delta y_{1} = 600 - 400$$$$\Delta y_{1} = 200$$[/tex]

and,  [tex]$\Delta y_{2}$[/tex], we subtract the 3rd y value from the 2nd y value[tex]$$y_{2} = 700$$ $$y_{1} = 600$$$$\Delta y_{2} = y_{2} - y_{1}$$$$\Delta y_{2} = 700 - 600$$$$\Delta y_{2} = 100$$[/tex]

To get [tex]$\Delta y_{3}$[/tex], we subtract the 4th y value from the 3rd y value.

[tex]$$y_{3} = 680$$ $$y_{2} = 700$$$$\Delta y_{3} = y_{3} - y_{2}$$$$\Delta y_{3} = 680 - 700$$$$\Delta y_{3} = -20$$[/tex]

Now let's substitute these values into the Newton's Forward Interpolation Formula;

[tex]$$y_{n} = y_{n-1} + \frac{(x-x_{n-1})}{h}(\Delta y)_{n-1} + \frac{(x-x_{n-1})(x-x_{n-2})}{2!h^{2}}(\Delta^{2}y)_{n-2} +[/tex] [tex]...+ \frac{(x-x_{n-1})(x-x_{n-2})...(x-x_{n-k+1})}{k!h^{k}}(\Delta^{k}y)_{n-k+1}$$[/tex]

where

x= 470

h= 10 (From the table)

x₀ = 500

y₀= 400

[tex]\\$\Delta y_{1} = 200$ \\$\Delta y_{2} = 100$ \\$\Delta y_{3} = -20$[/tex]

Now,[tex]$$y_{1} = y_{0} + \frac{(x-x_{0})}{h}\Delta y_{1}$$$$[/tex]

[tex]= 400 + \frac{(470 - 500)}{10}200$$$$[/tex]

[tex]= 360[/tex]

and, [tex]$$ $$y_{2} = y_{1} + \frac{(x-x_{1})}{h}\Delta y_{2}$$$$[/tex]

= [tex]= 360 + \frac{(470 - 510)}{10}100$$$$[/tex]

=[tex]710$$[/tex]

and, [tex]$$y_{3} = y_{2} + \frac{(x-x_{2})}{h}\Delta y_{3}$$$$y_{3} = 710 + \frac{(470 - 900)}{10}(-20)$$$$y_{3} = 584$$[/tex]

Therefore, the power of engine for 470 revolutions per minute is 584 hp.

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