What is 18x77 with algorithm

2=4x+y

y=2-4x

Александр Мазепов

Step-by-step explanation:

Step 1:  Solve for y

[tex]2 = 4x + y[/tex]

[tex]2 - 4x = 4x + y - 4x[/tex]

[tex]y = 2 - 4x[/tex]

Step 2:  Solve for x

[tex]2 - 2 - 4x = 0 - 2[/tex]

[tex]-4x / -4 = -2 / -4[/tex]

[tex]x = 1/2[/tex]

Step 3:  Solve for y

[tex]y = 2 - 4(0)[/tex]

[tex]y = 2[/tex]

Step 4:  Graph the equation

Graph the x-intercept, (1/2, 0), the y-intercept, (0, 2) and draw a line between them.  Look at the attached picture for the graph:

Answer:

Pvalue > 0.10

Step-by-step explanation:

Given the hypothesis :

H0 : β1 = β4 = 0

H1 : Atleast one of βj is not 0

F statistic = 1.482 ;

Degree of freedom = 2 and 21 ;

DFnumerator = 2

DFdenominator = 21

Using the Pvalue calculator from Fstatistic ;

Pvalue(1.482, 2, 21) = 0.24999 = 0.25

Hence, Pvalue for the test is 0.25

Pvalue > 0.10

Answer:

45

Step-by-step explanation:

The length of the arc is equal to the central angle it sees.

Step-by-step explanation:

here's the answer to your question

Answer:

your answer will be Option D

Step-by-step explanation:

log 10

Answer:

Given that the maximum age of the textbook is 7.9 years and the minimum age of the textbook is 1.3 years.

Using the range rule, the standard deviation is estimated as,

S≈maximum−minimum/4

=7.9−1.3/4

=1.65

The required value of the approximate standard deviation is 1.65.

The standard deviation of the data is 1.65.

Standard deviation is the measure of the deviation of the data from the mean.

The total college textbooks is 10

The oldest book is 7.9 years old

The newest book is 1.3 years old

The standard deviation of range is equal to one fourth of the difference of maximum to minimum.

The standard deviation = ( 7.9 - 1.3 ) /4 = 1.65

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First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

Let's apply the first derivative of this f(x) function.

[tex]f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\[/tex]

Now apply the derivative to that to get the second derivative

[tex]f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\[/tex]

We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

-----------------------------------

Let's compute dy/dx. We'll use f(x) as defined earlier.

[tex]y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\[/tex]

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

[tex]\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\[/tex]

If you need a refresher on the quotient rule, then

[tex]\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\[/tex]

where P and Q are functions of x.

-----------------------------------

This then means

[tex]\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\[/tex]

Note the cancellation of -(f ' (x))^2 with (f ' (x))^2

------------------------------------

Let's then replace f '' (x) with -p^2*f(x)

This allows us to form  ( f(x) )^2 in the numerator to cancel out with the denominator.

[tex]\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\[/tex]

So this concludes the proof that [tex]\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2 = 0\\\\[/tex] when [tex]y = \ln\left(\sin(px)+\cos(px)\right)\\\\[/tex]

Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.

Answer:

$1.25x

Step-by-step explanation:

Given :

Cost per pound = $1.25

Number of pounds of apple = x

The total cost of apple = (cost per pound * number of apple in pounds)

Hence,

Total cost of x pounds of apple is :

($1.25 * x)

= $1.25x

Answer:

0.01

Step-by-step explanation:

6/7% = 6÷7÷100 = 0.0085714286 round to the nearest hundredth = 0.01

Answer:

g(x) = x² + 2x + 1

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Functions

Step-by-step explanation:

Step 1: Define

Identify

g(x) = f(x + 1)

f(x) = x²

Step 2: Find

Answer:(

fx).(gx)=D. -40x^3+25x^2+45

Step-by-step explanation:

We know

[tex]\boxed{\sf Q=CV}[/tex]

[tex]\\ \large\sf\longmapsto C=\dfrac{Q}{V}[/tex]

[tex]\\ \large\sf\longmapsto C=\dfrac{40}{10}[/tex]

[tex]\\ \large\sf\longmapsto C=4\mu F[/tex]

Answer:

a) 10,025 m = 10km 25m = 10.025 km

b) 1,500 m = 1 km 500 m = 1.5 km

Answer:

a) 10025m = 10km 25m = 10.025km

b) 1500m = 1km 500m = 1.5km

Step-by-step explanation:

Concept:

Here, we need to know the idea of unit conversion.

Unit conversion is the conversion between different units of measurement for the same quantity.

1 km = 1000 m

Solve:

a)

b)

Hope this helps!! :)

Please let me know if you have any questions

Answer:

The ratio of the perimeters of the first nonagon to the second is 3.6 to 1.

Step-by-step explanation:

Given that a regular nonagon has an area of 90 square feet, and a similar nonagon has an area of 25 square feet, to determine what is the ratio of the perimeters of the first nonagon to the second, the following calculation must be performed:

25 = 1

90 = X

90/25 = X

3.6 = X

Therefore, the ratio of the perimeters of the first nonagon to the second is 3.6 to 1.

Answer:

0.9922 = 99.22% probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The production manager claims they have a mean life of 83 months with a variance of 81.

This means that [tex]\mu = 83, \sigma = \sqrt{81} = 9[/tex]

Sample of 146:

This means that [tex]n = 146, s = \frac{9}{\sqrt{146}}[/tex]

What is the probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors?

This is 1 subtracted by the p-value of Z when X = 81.2. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{81.2 - 83}{\frac{9}{\sqrt{146}}}[/tex]

[tex]Z = -2.42[/tex]

[tex]Z = -2.42[/tex] has a p-value of 0.0078.

1 - 0.0078 = 0.9922.

0.9922 = 99.22% probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors.

Assuming the car's speed [tex]\frac{630}{14}=45\mathrm{mph}[/tex] does not change, the car will travel [tex]45\cdot42=\boxed{1890}[/tex] miles.

Hope this helps :)

Answer:

Kindly check explanation

Step-by-step explanation:

The power of a test simply gives the probability of Rejecting the Null hypothesis, H0 in a statistical analysis given that the the alternative hypothesis, H1 for the study is true. Hence, the power of a test can be referred to as the probability of a true positive outcome in an experiment.

Using this definition, a power of 90% simply means that ; there is a 90% probability that the a Pvalue less Than the α - value of an experiment is obtained if there is truly a significant difference. Hence, a 90% chance of Rejecting the Null hypothesis if truly the alternative hypothesis is true.

Answer:

let 2 no. be x and y

x+y=31 ..... (1)

x-y=11 .........(2)

from (1)

x=31-y ..........(3)

putting (3) in (2)

31-y-y=11

-2y=11-31

-2y=-20

2y=20

y=10

Answer:

155 women must be randomly selected.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1 - 0.9}{2} = 0.05[/tex]

Now, we have to find z in the Z-table as such z has a p-value of [tex]1 - \alpha[/tex].

That is z with a pvalue of [tex]1 - 0.05 = 0.95[/tex], so Z = 1.645.

Now, find the margin of error M as such

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

The population standard deviation is known to be 28 lbs.

This means that [tex]\sigma = 28[/tex]

We want 90% confidence that the sample mean is within 3.7 lbs of the populations mean. How many women must be sampled?

This is n for which M = 3.7. So

[tex]M = z\frac{\sigma}{\sqrt{n}}[/tex]

[tex]3.7 = 1.645\frac{28}{\sqrt{n}}[/tex]

[tex]3.7\sqrt{n} = 1.645*28[/tex]

[tex]\sqrt{n} = \frac{1.645*28}{3.7}[/tex]

[tex](\sqrt{n})^2 = (\frac{1.645*28}{3.7})^2[/tex]

[tex]n = 154.97[/tex]

Rounding up:

155 women must be randomly selected.

Answer:

The proportion of bottles that will have more than 30 ounces dispensed into them is 0.9772 = 97.72%.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The amount dispensed has been observed to be approximately normally distributed with mean 32 ounces and standard deviation 1 ounce.

This means that [tex]\mu = 32, \sigma = 1[/tex]

Determine the proportion of bottles that will have more than 30 ounces dispensed into them.

This is 1 subtracted by the p-value of Z when X = 30, so:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{30 - 32}{1}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a p-value of 0.0228.

1 - 0.0228 = 0.9772

The proportion of bottles that will have more than 30 ounces dispensed into them is 0.9772 = 97.72%.

Answer:

-5≤x <1

Step-by-step explanation:

sqrt( x+5) / sqrt(1-x)

The numerator must be greater than zero since it is a square root

sqrt(x+5) ≥0

Square each side

x+5≥0

x≥-5

The denominator must be greater than zero (the denominator cannot be zero)

sqrt(1-x)> 0

Square each side

1-x > 0

1>x

Putting these together

-5≤x <1

The average weekly allowance of a 5th grade student as calculated using direct variation with the information provided by Fidelity Investment Vision Magazine is 5.367 dollars per week.

The question given is a direct variation problem:

Let:

• Average weekly allowance = [tex]a[/tex]

• Grade level = [tex]g[/tex]

If Average weekly allowance varies directly as grade level , then , then the direct variation between the variables can be expressed as :

[tex]a = k * g[/tex]

Where ,  [tex]k[/tex] = constant of proportionality

We can obtain the value of k from the given values of a and g

[tex]9.66 = k * 9\\9.66 = 9k\\k = 9.66/9[/tex]

Our equation becomes:

[tex]a = (9.66/9) * g[/tex]

[tex]a = (9.66/9) * 5\\a = 5.367[/tex] (rounded to 3 decimal places)

Hence, using proportional relationship, the average weekly allowance for a 5th grade student is [tex]5.367[/tex] per week

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Answer:

1

Step-by-step explanation:

Cosine is a trigonometric function that is represented by adjacent divided by the hypotenuse. The side adjacent to angle A is AC and the hypotenuse is AB, so we can say cos(A) = [tex]\frac{AC}{AB}[/tex]. We can do the same for angle B. The side adjacent to it is BC, and the hypotenuse is again AB. So, we can say

cos(B) = [tex]\frac{BC}{AB}[/tex]. We are solving for [tex]\frac{cosA}{cosB}[/tex], so we can substitute the value of those two and solve:

[tex]\frac{\frac{AC}{AB}}{\frac{BC}{AB} }[/tex]

[tex]\frac{AC}{AB} * \frac{AB}{BC} = \frac{AC}{BC}[/tex]

AC is given to be 3 and BC is also 3, so [tex]\frac{AC}{BC}[/tex] is [tex]\frac{3}{3}[/tex] which is just 1.

Answer:

Reject H0 and conclude that volume filled by old machine varies more than volume filled by new machine

Step-by-step explanation:

Given the data:

Old machine: 8.2, 8.0, 7.9, 7.9, 8.5, 7.9, 8.1,8.1, 8.2, 7.9

New machine: 8.0, 8.1, 8.0, 8.1, 7.9, 8.0, 7.9, 8.0, 8.1

To test if volume filled by old machine varies more than volume filled by new machine :

Hypothesis :

H0 : s1² = s2²

H1 : s1² > s2²

Using calculator :

Sample size, n and variance of each machine is :

Old machine :

s1² = 0.37889

n = 10

New machine :

s2² = 0.006111

n = 9

Using the Ftest :

Ftest statistic = larger sample variance / smaller sample variance

Ftest statistic = 0.37889 / 0.006111

Ftest statistic = 62.001

Decision region :

Reject H0 ; If Test statistic > Critical value

The FCritical value at 0.05

DFnumerator = 10 - 1 = 9

DFdenominator = 9 - 1 = 8

Fcritical(0.05, 9, 8) = 3.388

Since 62 > 3.388 ; Reject H0 and conclude that volume filled by old machine varies more than volume filled by new machine

Answer:

4 ft

Step-by-step explanation:

I guess, the meaning is the largest marbles, so that we can pave the whole floor without cutting any marbles and leaving empty spots.

so, 20×16 ft²

we can have marbles 1/2 ft long. and it all fits well : 40×32 marbles.

we can have them 1 ft long, and it all fits well : 20×16 marbles.

we can have them 2ft long, and it still fits well : 10×8 marbles.

and so on.

so, actually, we are looking for the greatest common divisor (GCD) of 20 and 16. and that gives us the maximum length of a single marble to fulfill the requirement.

let's go for the prime factors starting with 2

20/2 = 10

10/2 = 5

5/3 fits not work

5/5 = 1 done

so, 20 = 2²×3⁰×5¹

16/2 = 8

8/2 = 4

4/2 = 2

2/2 = 1 done

16 = 2⁴

so, for the GCD I can only use powers of 2 (the only prime factors both numbers have in common).

and we have to use the smaller power of 2, which is 2, so, the GCD is 2² = 4

=>

the maximum length of the squared marbles is 4 ft.

that would pave the floor with 5×4 marbles completely.

Answer: B. This function has no intercept. I think B is the correct answer.

Answer:

x=3

Step-by-step explanation:

Find the radius and use Pythagoras on the right side