to find the area between two z-scores, we look up the area to the ____ of each value and subtract the smaller area from the larger area.

The percent increase in the population over the initial population after 7 hours is approximately 40.7%.

To solve this problem, we can use the formula for exponential growth:

P(t) = P0(1 + r)^t

Where P(t) is the population after t hours, P0 is the initial population, r is the growth rate as a decimal (in this case, 0.05), and t is the time in hours.

Plugging in the given values, we get:

P(7) = P0(1 + 0.05)^7

To find the percent increase in population over the initial population, we can subtract the initial population from the final population, divide by the initial population, and then multiply by 100:

Percent increase = [(P(7) - P0)/P0] x 100

Simplifying this expression using the formula for exponential growth, we get:

Percent increase = [(1 + 0.05)^7 - 1] x 100

Calculating this expression using a calculator or spreadsheet, we get:

Percent increase ≈ 40.7%

Therefore, the percent increase in the population over the initial population after 7 hours is approximately 40.7%.

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Answer: C

Step-by-step explanation:

area of rectangle = (5--2) × (2--2) = 7 × 4 = 28

area of triangle = ((12-5) × (6--6)) ÷ 2 = (7 × 12) ÷ 2 = 84 ÷ 2 = 42

total area = 28 + 42 = 70

The equation that Mai could have written, using the Fermi process is A. ( 4x 10 ⁵ ) / ( 4x 10 ⁻⁴ ) = 1 x 10 ⁹.

A technique initiated by Fermi is used for predicting a rough figure or estimation of something, typically with minuscule or no details about the particulars of the thing being predicted.

The volume of the pack of gums would be:

= 1 / 5 x 1 / 10 x 1 / 50

= 0. 0004

= 4x 10 ⁻⁴

The volume of the gymnasium would be:

= 100 x 80 x 50

= 400, 000

= 4x 10 ⁵

So this can be written as:

( 4x 10 ⁵ ) / ( 4x 10 ⁻⁴ ) = 1 x 10⁹ .

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The sample size of flights required is 159.

Sample size determination is the process of calculating the number of individuals or items that need to be included in a sample to obtain statistically significant results in a study.

The sample size is determined based on the population size, the level of confidence desired, the margin of error, and the expected variability in the data.

We can use the formula for the margin of error of a confidence interval:

Margin of error = z × (standard deviation / √(sample size))

Here we have

An initial study of US domestic flights produced 81 as the standard deviation of the flight times (in minutes).

The confidence level is 98%

We want the margin of error to be 15 minutes,

So we can rearrange the formula to solve for the sample size:

=> sample size = (z × standard deviation / margin of error)²

Substituting in the given values, we get:

=> sample size = (2.33 × 81 / 15)² = 158.3

Rounding up to the nearest whole number, we need a sample size of at least 159 flights.  

Therefore,

The sample size of flights required is 159.

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A)  The average velocity of the particle is 4.5 feet per second

B)  The particle reaches its average velocity at time c is 5.25 seconds.

The average velocity of a particle on the interval [1,3] is given by the formula:

The change in displacement is given by

The change in time is 3 - 1 = 2 seconds.

Therefore, the average velocity of the particle is:

To find the time c when the particle reaches its average velocity, we need to solve the equation:

where s(t) = -t² + 5t.

Substituting s(t) into the equation, we get:

Simplifying the equation, we get:

Solving for c, we get:

Therefore, the particle reaches its average velocity at time c = 5.25 seconds.

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The survival rates among passengers in other classes, cannot definitively determine whether survival and ticket class are independent.

No, because the probability of survival and the probability of survival given a first-class passenger are not the same. C

To determine whether survival and ticket class are independent, to compare the probability of survival among all passengers to the probability of survival among first-class passengers.

The probability of survival is the same for all passengers, regardless of their ticket class, then we can conclude that survival and ticket class are independent.

The probability of survival varies depending on the ticket class, then we cannot conclude that they are independent.

The probability of survival among all passengers is 0.268.

The probability of survival among first-class passengers is also 0.268. This does not necessarily mean that survival and ticket class are independent.

To determine whether they are independent, we need to compare the probability of survival among all other ticket classes as well.

If the probability of survival is the same across all ticket classes, then we can conclude that survival and ticket class are independent.

The probability of survival varies across different ticket classes, then we cannot conclude that they are independent.

The survival rates among passengers in other classes, cannot definitively determine whether survival and ticket class are independent.

The fact that the survival rate among first-class passengers is the same as the overall survival rate does not prove independence.

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The area between the three curves is approximately 1.175 square units.

By counting the number of squares on a piece of paper with grids (square shaped), and using basic formulas, it is possible to determine the area of shapes like quadrilaterals and circles, which are 2D shapes.

To find the area between the curves, we first need to determine the points of intersection.

Setting the first two equations equal to each other gives:

3/x = 12x

x² = 1/4

x = 1/2

Substituting x = 1/2 into either of the equations gives y = 6, so the first two curves intersect at (1/2, 6).

Setting the second and third equations equal to each other gives:

12x = 1/12x

x² = 1/144

x = 1/12

Substituting x = 1/12 into either of the equations gives y = 1, so the second and third curves intersect at (1/12, 1).

Thus, we can see that the region bounded by the curves is composed of two parts, which we can find separately and then add together.

First, we find the area between y = 3/x and y = 12x, which is bounded by x = 1/12 and x = 1/2. To find the area, we integrate the difference between the two functions with respect to x:

A1 = ∫(1/12 to 1/2) (12x - 3/x) dx

= [6x² - 3ln(x)] from x = 1/12 to x = 1/2

= [3/8 - 3ln(1/12)] - [1/144 - 3ln(1/2)]

= 3/8 + 3ln(12) - 1/144

Next, we find the area between y = 12x and y = 1/12x, which is bounded by x = 1/12 and x = 1/2. To find the area, we integrate the difference between the two functions with respect to x:

A2 = ∫(1/12 to 1/2) (3/x - 1/12x) dx

= [3ln(x) - (1/24)x²] from x = 1/12 to x = 1/2

= [3ln(1/2) - (1/4)(1/12)²] - [3ln(1/12) - (1/4)(1/2)²]

= 3ln(2) - 1/144 - 3ln(12) + 1/16

= 3ln(2) - 3ln(12) + 1/16 - 1/144

Now, we can find the total area by adding the two areas:

A = A1 + A2

= 3/8 + 3ln(12) - 1/144 + 3ln(2) - 3ln(12) + 1/16 - 1/144

= 1/16 + 3ln(2)

Therefore, the area between the three curves is approximately 1.175 square units.

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a) The most appropriate statistical test, for testing the random sample mean of two samples is test for two independent means. So, option(vo) is right one.

b) The appropriate hypotheses for this is

[tex]H_0 : \mu_1 = \mu_2 [/tex]

[tex]H_a : \mu_1 > \mu_2 [/tex].

We have a random samples survey of Oakland apartments and Shadyside apartments. Claim is that overall mean monthly rent of apartments in Shadyside is higher than in Oakland.

a) We determine the most appropriate test : From the information, we consider a random sample of Shadyside apartments. In this situation, observe that there are two samples of monthly rents in apartments in Shadyside and Oakland cities and also compares the mean rent of apartments in Shadyside and Oakland cities. Therefore, the researcher uses two sample mean test. Hence, the correct option is (vi).

b) Now, Let the sample means for monthly rents in apartments in Shadyside and Oakland cities be [tex] \mu_1 and \mu_2 [/tex] respectively. So, the appropriate hypotheses, using the appropriate parameter that is null and alternative hypothesis are [tex]H_0 : \mu_1 = \mu_2 [/tex]

[tex]H_a : \mu_1 > \mu_2 [/tex]. Hence, required value is occured.

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a) Simple events in the sample space: {B}, {GB}, {GGB}, {GGGB}, {GGGG}.

b)Probability of each simple event: {B} = 0.5, {GB} = 0.25, {GGB} = 0.125, {GGGB} = 0.0625, {GGGG} = 0.0625.

c) Probability distribution function for X: P(X=1) = 0.5, P(X=2) = 0.25, P(X=3) = 0.1875, P(X=4) = 0.0625.

d)The graph of the probability distribution function for X would have a bar at X=1 with height 0.5, a bar at X=2 with height 0.25, a bar at X=3 with height 0.1875, and a bar at X=4 with height 0.0625. The graph would have a right-to-left bias.

a) The sample space's simple events are B, GB, GGB, GGGB, and GGGG.

b) The probability of each simple event can be calculated by multiplying the probabilities of having a boy or a girl for each birth until the woman stops having children. For example, the probability of {GB} is 0.5*0.5 = 0.25, since the woman must have a boy on the first birth and a girl on the second birth. The probabilities of the other simple events are: {B} = 0.5, {GGB} = 0.125, {GGGB} = 0.0625, and {GGGG} = 0.0625.

c) The probability distribution function for X can be found by adding up the probabilities of all the simple events that result in X children. For example, P(X=1) = P({B}) = 0.5, P(X=2) = P({GB}) = 0.25, P(X=3) = P({GGB, GGGB}) = 0.125 + 0.0625 = 0.1875, and P(X=4) = P({GGGG}) = 0.0625.

d) The probability distribution function for X can be visualized using a bar graph, where the height of each bar represents the probability of having a certain number of children.

The graph would have a bar at X=1 with height 0.5, a bar at X=2 with height 0.25, a bar at X=3 with height 0.1875, and a bar at X=4 with height 0.0625. The graph would be skewed to the right, since the probability of having fewer children is higher than the probability of having more children.

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The probabilities of Sue catching the bus TO the shops, catching the bus FROM the shops, walking to the shops, and walking from the shops are 0.4, 0.7, 0.6, and 0.3, respectively, given that she either catches the bus or walks when visiting the shops.

Let's denote the event of Sue catching the bus TO the shops as A and the event of her catching the bus FROM the shops as B. Then, we can use the following probabilities:

P(A) = 0.4

P(B) = 0.7

Since Sue either catches the bus or walks, these two events are mutually exclusive and exhaustive. Therefore, the probability of her walking to the shops is:

P(not A) = 1 - P(A) = 1 - 0.4 = 0.6

Similarly, the probability of her walking from the shops is:

P(not B) = 1 - P(B) = 1 - 0.7 = 0.3

We can also use the law of total probability to find the probability of Sue catching the bus:

P(bus) = P(A) + P(B) = 0.4 + 0.7 = 1.1

This value is greater than 1, which is not possible since probabilities cannot be greater than 1. This means that there is an error in the given probabilities. However, we can still use the above calculations for the given probabilities to determine the probabilities of walking and catching the bus.

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The complete question is :

What are the probabilities of Sue catching the bus TO the shops, catching the bus FROM the shops, walking to the shops, and walking from the shops, if the probability of Sue catching the bus TO the shops is 0.4 and the probability of her catching the bus FROM the shops is 0.7, and it is known that she either catches the bus or walks when visiting the shops?

The term that best describes the investment is $1,250 invested at 4% compounded interest. The Option A is correct.

We will use compound interest: A = P(1 + r/n)^(nt) formula to get the annual total investment. In this case, P = $1,250, r = 0.04, n = 1 and t = 1.

Plugging the values, we get:

A = 1250(1 + 0.04/1)^(1*1)

A = 1250(1.04)

A = $1,300

Therefore, term that best describes the investment is $1,250 invested at 4% compounded interest.

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Country B has a higher labor productivity than country A. This means that, on average, each worker in country B is producing more final goods per hour worked compared to each worker in country A.
Based on the given information, we can calculate the labor productivity of both countries.

Country A:
- Labor force = 800
- Hours worked per day = 8
- Total labor hours per day = 800 x 8 = 6,400
- Total final goods produced per day = 128,000
- Labor productivity = 128,000 / 6,400 = 20

Country B:
- Labor force = 1,800
- Hours worked per day = 6
- Total labor hours per day = 1,800 x 6 = 10,800
- Total final goods produced per day = 270,000
- Labor productivity = 270,000 / 10,800 = 25

Therefore, country B has a higher labor productivity than country A. This means that, on average, each worker in country B is producing more final goods per hour worked compared to each worker in country A.

Based on the provided information, Country A has a population of 1,000 with 800 people working 8 hours a day, producing 128,000 final goods. Country B has a population of 2,000, with 1,800 people working 6 hours a day, resulting in 270,000 final goods.

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The numbers are located on the apposite side of 0 from point A are -3, -5. So, correct options are A, B.

Point A is located at the opposite of -7 on the number line. The opposite of a number is the value that gives a sum of zero when added to the original number. Therefore, the opposite of -7 is 7.

To find the numbers located on the opposite side of 0 from point A, we need to determine the sign of 7. Since 7 is a positive number, the numbers on the opposite side of 0 will be negative numbers.

Thus, the numbers located on the opposite side of 0 from point A are all negative numbers. This includes all numbers less than zero, such as -1, -2, -3, -4, -5, -6, -7, -8, -9, and so on.

Therefore, the correct options are A, B.

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Complete question is:

On a number line,point A is located at the apposite of -7. Which numbers are located on the apposite side of 0 from point A? Mark all that apply

A) -3

B) -5

C) 3

D) 5

Based on the given algebra tile configuration, Adi correctly represented the product (negative 2 x minus 1)(2 x minus 1). So, correct option is A.

In the Factor 1 spot, Adi used 4 tiles, 2 of which were labeled negative x and 2 labeled negative. This correctly represents the factor negative 2 x minus 1.

In the Factor 2 spot, Adi used 3 tiles, 2 of which were labeled positive x and 1 labeled negative. This correctly represents the factor 2 x minus 1.

In the Product spot, Adi used 12 tiles, with 4 labeled negative x squared, 4 labeled negative x, 2 labeled positive x, and 2 labeled positive. These labels correctly represent the terms obtained by multiplying the terms in the Factor 1 spot and the Factor 2 spot.

Therefore, it can be concluded that Adi used the algebra tiles correctly to represent the product (negative 2 x minus 1)(2 x minus 1).

So, correct option is A.

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The score in the 4th test is 80 and the median of the hours worked is 6

Here, we have

Scores = 92, 88, 76

Mean = 84

So, we have

mean = sum/count

This gives

(92 + 88 + 76 + score)/4 = 84

Evaluate

score = 4 * 84 - (92 + 88 + 76)

So, we have

score = 80

Here, we have

Hours = 8, 6, 8, 6, 4

Sort in ascending order

So, we have

Hours = 4, 6, 6, 8, 8

The median of the hours worked is the middle value

So, we have

median = 6

hence, the median is 6

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The largest possible value of the greatest common divisor (GCD0 is 3.

To find the largest possible value of the greatest common divisor (GCD) of 2m and 3n, we need to first find the prime factorization of 2m and 3n.

We can start by factoring out 3 from the expression for m:

Therefore, 2m = 6(8n + 17).

Next, we can factor 3n as 3 times some integer k.

Now, the prime factorization of 2m is 2 × 3 × (8n + 17), and the prime factorization of 3n is 3 × k.

Since 2 and 3 have no common factors, the GCD of 2m and 3n will be equal to the GCD of (8n + 17) and k.

To maximize the GCD, we want to maximize (8n + 17) and k separately. The largest possible value for (8n + 17) occurs when n = 3, which gives us (8n + 17) = 41. For k, any value greater than or equal to 1 will work.

Therefore, the largest possible value of the GCD of 2m and 3n is GCD(2m, 3n) = GCD(6 × 41, 3k) = 3 × GCD(82, k), where k is any integer greater than or equal to 1. So the largest possible value of the GCD is 3.

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The frequency for the number of computer science students enrolled in Stat 155 is 127, and the relative frequency is approximately 0.552 (to 3 decimal places).

In Stat 155, there are 230 students enrolled, and 127 of them are majoring in computer science. The frequency for the number of computer science students enrolled in Stat 155 is 127. To find the relative frequency, divide the frequency by the total number of students:

Relative frequency = (Number of computer science students) / (Total number of students)
Relative frequency = 127 / 230 ≈ 0.552

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Step-by-step explanation:

Every year you get 3 % of 1800 pounds added to the account

  what is 3% of 1800 ?         3 % is .03 in decimal form

     1800 * .03 = 54  pounds interest per year

in TWO years you will have   54 * 2   added to 1800 = 1908 pounds

Answer:

A). increase by 54  per year.

B). After two years the account has 1908

Step-by-step explanation:

A). 800 * 3% * 1 = 54

B). 1800 + 1800 * 3% * 2  = 1908

Hope this helps

Answer:

50%

Step-by-step explanation:

To answer this question, we can use the formula for the probability of an event:

P(event)=total number of outcomes number of favorable outcomes​

In this case, the event is selecting a yellow T-shirt, so the number of favorable outcomes is 8 (the number of yellow T-shirts). The total number of outcomes is 16 (the number of T-shirts). Therefore, the probability is:

P(yellow)=168​=21​

This means that the probability of selecting a yellow T-shirt is one half or 0.5. We can also write this as a percentage: 50%.

Yes, we can fully describe the density curve for a normal distribution in terms of just the mean (μ) and standard deviation (σ).

The normal distribution is a symmetric, bell-shaped curve that is completely determined by its mean and standard deviation.

The mean (μ) determines the center or peak of the curve, and the standard deviation (σ) determines the spread or width of the curve. Specifically, the normal distribution has the following properties:

The mean, median, and mode of the distribution are all equal and located at the center of the curve.

The total area under the curve is equal to 1, which means that the curve represents the probability density function for all possible values of the random variable.

The curve is symmetric around the mean, with half of the area under the curve to the left of the mean and half to the right.

The standard deviation controls the width of the curve, with larger standard deviations resulting in wider, flatter curves and smaller standard deviations resulting in narrower, taller curves.

The mean and standard deviation of a normal distribution, we can easily calculate the probabilities associated with specific values or ranges of values using the properties of the curve.

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The polynomials span P₂ implies any polynomial of degree 2 written as linear combination of these polynomials.

To determine whether the given polynomials span P₂,

Check whether any polynomial of degree 2 can be written as a linear combination of these polynomials.

Let us consider a general polynomial of degree 2.

p(x) = ax² + bx + c

We need to find coefficients k₁, k₂, k₃, and k₄ such that.

p(x) = k₁(1-x+2x²) + k₂(3+x) + k₃(5-x+4x²) + k₄(-2-2x+2x²)

Expanding the right side and collecting like terms, we get,

p(x) = (2k₁+4k₃+2k₄)x² + (-k₁-k₂+k₃-2k₄)x + (k₁+3k₂+5k₃-2k₄+1)

This equation must hold for any value of x.

Equate the coefficients of the powers of x on both sides,

2k₁ + 4k₃ + 2k₄ = a

-k₁ - k₂ + k₃ - 2k₄ = b

k₁ + 3k₂ + 5k₃ - 2k₄ + 1 = c

Solve this system of linear equations for k₁, k₂, k₃, and k₄.

Write this in matrix form as,

[tex]\left[\begin{array}{cccc}2&0&4&2\\-1&-1&1&-2\\1&3&5&-2\end{array}\right][/tex][tex]\left[\begin{array}{ccc}k_{1} \\k_{2}\\k_{3}\end{array}\right][/tex]  [tex]= \left[\begin{array}{ccc}a \\b\\c-1\end{array}\right][/tex]

Solve this system using Gaussian elimination or other methods.

However, a simpler way to check whether the polynomials span P₂ is to check whether the matrix of coefficients is invertible.

If the matrix is invertible, then there is a unique solution for any value of a, b, and c.

If the matrix is not invertible,

Then there are some values of a, b, and c for which there is no solution.

And the polynomials do not span P₂.

To check whether the matrix is invertible, compute its determinant,

[tex]\left|\begin{array}{cccc}2&0&4&2\\-1&-1&1&-2\\1&3&5&-2\end{array}\right|[/tex]

= 12

Since the determinant is non-zero,

The matrix is invertible, and the polynomials span P₂.

Therefore, matrix is invertible implies polynomials span P₂, any polynomial of degree 2 written as linear combination of these polynomials.

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The above question is incomplete, the complete question is:

Determine whether the following polynomials span P₂ (polynomial of degree 2):

p₁=1−x+2x²,

p₂=3+x,

p₃=5−x+4x²,

p₄=−2−2x+2x²

The probability that a point within the white circle lies on the red circle is P = 0.44

The probability will be given by the quotient between the area of the red square and the area of the circle.

On the diagram we can only see two squares, so Im assuming that it should say "square" instead of circle.

The area of the large square is:

A = 3*3 = 9

The area of the red square is:

A' = 2*2 = 4

Then the probability is:

P = 4/9 = 0.44

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To calculate the p-value, we need to use a one-tailed t-test since we're interested in the probability of getting a sample mean greater than 12.10 oz.

First, we need to calculate the t-statistic:

t = (sample mean - hypothesized mean) / (standard deviation / sqrt(sample size))

t = (12.15 - 12.10) / (0.05 / sqrt(25))

t = 3.1623

Next, we need to find the degrees of freedom, which is the sample size minus one:

df = 25 - 1 = 24

Using a t-distribution table with 24 degrees of freedom and a significance level of 0.01, we find that the critical value is 2.492.

Since the calculated t-value (3.1623) is greater than the critical value (2.492), we can reject the null hypothesis and conclude that there is evidence that the mean volume is greater than 12.10 oz.

To find the p-value, we need to calculate the probability of getting a t-value greater than 3.1623 with 24 degrees of freedom:

p-value = P(t > 3.1623) = 0.0028 (calculated using a t-distribution table or software)

Therefore, the p-value is 0.0028, which is less than the level of significance (0.01), indicating strong evidence against the null hypothesis.

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We need a resistor that can handle at least 0.075 W of power.

To calculate the resistance needed, we can use Ohm's Law, which states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them. Mathematically, Ohm's Law can be expressed as:

V = IR

Where V is the voltage, I is the current, and R is the resistance. Rearranging this equation, we can solve for the resistance:

R = V/I

Plugging in the values given, we get:

R = 2.5 V / 0.3 A = 8.33 ohms

Therefore, we need a resistance of 8.33 ohms to connect the 2.5 V, 0.3 A light bulb to the flat battery. One way to connect the light bulb is to use a resistor in series with the bulb.

It's important to choose a resistor that can handle the power dissipation, which is given by:

P = IV = I²R = V²/R

In this case, the power dissipation is:

P = (0.3 A)² x 8.33 ohms = 0.075 W

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Complete Question:

We want to connect a 2.5 V, 0.3 A light bulb to a flat battery. How much resistance and how do we need to connect to the light bulb?

Answer: 361

Step-by-step explanation:

The solution to the system of equations is: (x, y, z) = (-5/9, 19/9, 5/3)

What is algebra?

Algebra is a branch of mathematics that deals with mathematical operations and symbols used to represent numbers and quantities in equations and formulas. It involves the study of variables, expressions, equations, and functions.

Using Gaussian elimination, we can write the augmented matrix of the system:

\begin{pmatrix}1 & 1 & -2 & 2\2 & -1 & -1 & 0\6 & 3 & 4 & 19\end{pmatrix}

We can use elementary row operations to transform this matrix into row echelon form:

R2 = R2 - 2R1

R3 = R3 - 6R1

\begin{pmatrix}1 & 1 & -2 & 2\0 & -3 & 3 & -4\0 & -3 & 16 & 7\end{pmatrix}

Now we can use elementary row operations to transform this matrix into reduced row echelon form:

R2 = -1/3R2

R3 = R3 - R2

\begin{pmatrix}1 & 1 & -2 & 2\0 & 1 & -1 & 4/3\0 & 0 & 1 & 5/3\end{pmatrix}

Finally, we can use back substitution to find the solution:

z = 5/3

y - z = 4/3, y = 4/3 + z = 19/9

x + y - 2z = 2, x = 2 + 3z - y = -5/9

Therefore, the solution to the system of equations is:

(x, y, z) = (-5/9, 19/9, 5/3)

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The appropriate statistical test for this scenario would be a paired t-test for means, since the same group of individuals were tested twice under two different conditions (using word processor A and B).

The null hypothesis would be that there is no significant difference in typing speed between the two word processors, while the alternative hypothesis would be that there is a significant difference. To conduct the test, the differences between the typing speeds for each individual would be calculated (speed when using word processor A minus speed when using word processor B), and the mean and standard deviation of these differences would be calculated.

Then, a t-statistic would be calculated using the formula (mean difference / standard error of the mean difference), and the p-value would be determined based on the t-distribution with degrees of freedom equal to the sample size minus 1. If the p-value is less than the chosen level of significance (usually 0.05), then we reject the null hypothesis and conclude that there is a significant difference in typing speed between the two-word processors.

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If the construction cones gets filled with water due to heavy rain, then the volume of water that filled one-cone is 823.2 in³.

The "Volume" is defined as measure of amount of space occupied by a three-dimensional object. It is expressed in cubic units,

The volume of a cone can be calculated using the formula : V = (1/3)πr²h,

where V denotes volume, 'r" = radius, "h" = height, and π ≈ 3.14;

The diameter of the cone is 11 inches, so radius is = 11/2 = 5.5 inches;

Substituting the values,

We get,

⇒ V = (1/3) × π × (5.5)² × (26);

⇒ V ≈ 823.2 cubic inches,

Since the cone is filled with water, the volume of the water is equal to the volume of the cone.

Therefore, volume of water that filled one cone is approximately 823.2 cubic inches.

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a) The points[tex](x, y) = (-1, 2)[/tex] are the equilibrium solutions

b) The solutions are decreasing since [tex](y-2)[/tex] is negative and [tex](x+1)[/tex] is positive.

c) The solution curves concave up if y'' is positive, and concave down if y'' is negative.

a)  Either [tex]y = 2 or x = -1[/tex]is required for this equation to be true. Therefore, the points[tex](x, y) = (-1, 2)[/tex] are the equilibrium solutions.

We must set [tex]y = 0[/tex] and solve for y in order to find the equilibrium solutions. So:

[tex](y-2)(x+1) = 0[/tex]

b) We need to look at the sign of y' in various areas of the xy-plane to figure out where the solutions are rising or decreasing. The solutions are growing if y' is positive; they are shrinking if y' is negative.

Since [tex](y-2)[/tex] and [tex](x+1)[/tex]are both negative, y' is positive and the solutions are increasing if[tex]x -1[/tex]and [tex]y 2.[/tex] When [tex]x > -1[/tex]and [tex]y 2, (y-2)[/tex] becomes negative and (x+1) becomes positive, indicating that y' is negative and the solutions are getting smaller. If [tex]y > 2[/tex], then y' is positive and the solutions are getting bigger because [tex](y-2)[/tex] and [tex](x+1)[/tex] are both positive. for x is greater than [tex]-1[/tex] and y is greater than [tex]2[/tex], the solutions are decreasing since [tex](y-2)[/tex] is negative and [tex](x+1)[/tex] is positive. This is the case for [tex]y 2.[/tex]

The expression for y' shows that when [tex](y-2)[/tex] and [tex](x+1)[/tex]have the same sign, and when they have the opposite sign, respectively, it will be positive.

c) We need to look at the sign of y'' in various areas of the xy-plane to figure out where the solution curves are concave up-concave down. By taking the derivative of y', we may find y'':

[tex]y'' = (y-2) - 2(x+1)[/tex]

The solution curves concave up if y'' is positive, and concave down if y'' is negative.

We may deduce that y'' is positive when[tex]y > 2 + 2(x+1)[/tex] and negative when [tex]y 2 + 2(x+1)[/tex] from the expression for y''. As a result, when the solution curves are above the line[tex]y = 2 + 2(x+1)[/tex], they are concave up, and when they are below it, they are concave down.

a) [tex]y > 2 + 2(x+1)[/tex]

b) [tex]y 2 + 2(x+1)[/tex]

c)[tex]y = 2 + 2(x+1)[/tex]

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Complete Question:

Let y' =(y-2)(x+1). a) Determine all equilibrium solutions. b) Determine the region in the xy - plane where the solutions are increasing, and where the solutions are decreasing. c) Determine the regions in the xy - plane where the solution curves are concave up, and determine those regions where they are concave down. Solve the following differential equations.

a) dy + 2xy2 = 0 doc ? =

b) x - y = 2x?y, y(i)=1 y *1) b dy dx

Answer:

66 coins.

Step-by-step explanation:

There are 4 quarters in a year, so 2 years is 8 quarters.

Since Janice adds 5 coins every 3 months, in one year (or 4 quarters), she will add:

5 coins/3 months x 4 quarters = 20 coins

So in 2 years (8 quarters), she will add:

20 coins/year x 2 years = 40 coins

Therefore, after 2 years, the total number of coins in Janice's collection will be:

26 + 40 = 66 coins.