solve for x in the equation 6(–5)=54 6 ( x – 5 ) = 54 .

Answer:

a) The mean of the sampling distribution of means μₓ = μ = 20

b) The standard deviation of the sample σₓ⁻ = 0.5

Step-by-step explanation:

Explanation:-

Given sample size 'n' =100

Given  mean of the Population 'μ' = 20

Given standard deviation of population 'σ' = 5

a) The mean of the sampling distribution of means μₓ = μ

                                                                                     μₓ = 20

b) The standard deviation of the sample σₓ⁻   =    [tex]\frac{S.D}{\sqrt{n} }[/tex]

    Given standard deviation of population 'σ' = 5

                                                                              = [tex]\frac{5}{\sqrt{100} } = 0.5[/tex]

Final answer:

The mean of the distribution of the sample means drawn from a population with a mean of 20 and a standard deviation of 5, using samples of size 100, is 20. The standard deviation of this distribution, also known as the standard error, is 0.5.

Explanation:

The question involves understanding the concept of the distribution of sample means, also known as the sampling distribution. When samples of size 100 are drawn from a population with a mean (μ) of 20 and a standard deviation (σ) of 5, the mean of the distribution of the sample means will be the same as the population mean, which is 20. However, the standard deviation of the distribution of sample means, known as the standard error (SE), will be the population standard deviation divided by the square root of the sample size (√n), which in this case is 5/√100 = 0.5.

Therefore, the mean of the distribution of the sample means is 20 and the standard deviation (standard error) of this distribution is 0.5. This is based on the central limit theorem which states that, for a sufficiently large sample size, the sampling distribution of the sample mean will be approximately normally distributed regardless of the population’s distribution, with these exact parameters of mean and standard deviation.

Answer:

use 6cm twice and that adds up to 12cm .six is the minimal possible number

A ruler designed to measure lengths from one unit to twelve units with the minimum number of marks would only have marks at every unit. This results in only eleven markings (from 1 to 11), as the 0 marks the start and the twelfth unit marks the end.

The subject of the student's question pertains to the design of a ruler with the minimum possible number of marks that can measure lengths from one to twelve units. To achieve this, the ruler should only be marked at every unit. This way, there would be only eleven marks (1-11), as the 0 marks the start of the ruler and the twelfth unit is considered the end of the ruler but does not require a mark. Therefore, only the spaces between these numbers need to be marked to enable measurement all the way up to twelve units.

In practice, such a ruler is less precise than a ruler with more marks as the latter can measure smaller lengths more accurately. However, the task demands the bare minimum for a measure from 1 to 12, thus we have eleven marks.

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Answer:

The answer is (A)

Answer:

The answer is A.

Step-by-step explanation:

This is because the plus sign after the first mixed number gets multiplied to the second mixed number. When a plus sign gets multiplied to something that is negative already, then the answer is negative, so the equation will still stay the same and be -3 1/8 - 2 1/2.

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Answer:

7

Step-by-step explanation:

Trial and error.  Start with 1 and go to 9

The digit A in the numeral 3AA1, which is divisible by 9, represents the number 7. This is deduced by the rule that a number is divisible by 9 if the sum of its digits is also divisible by 9.

To find the digit that A represents in the four-digit numeral 3AA1 that is divisible by 9, we must remember that a number is divisible by 9 if the sum of its digits is divisible by 9. Let's denote A as a single digit, and since we know that 3 and 1 are already part of the sum, the equation we need to solve is:

3 + A + A + 1 = 3 + 2A + 1

Simplifying further:

2A + 4 must be divisible by 9. Since A is a digit, it can only be one of the following numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. We can now test these values:

For A = 0, the sum is 4 (not divisible by 9)

For A = 1, the sum is 6 (not divisible by 9)

For A = 2, the sum is 8 (not divisible by 9)

For A = 3, the sum is 10 (not divisible by 9)

For A = 4, the sum is 12 (not divisible by 9)

For A = 5, the sum is 14 (not divisible by 9)

For A = 6, the sum is 16 (not divisible by 9)

For A = 7, the sum is 18 (divisible by 9)

For A = 8, the sum is 20 (not divisible by 9)

For A = 9, the sum is 22 (not divisible by 9)

The only value for A that makes the sum divisible by 9 is 7. Therefore, A represents the digit 7.

Answer:

The risk consequence of both activities is 1.6 months.

Step-by-step explanation:

We can define a randome variable D: total delay time for the project, and calculate its expected value.

This would be the risk consequence of both activities.

The expected delay time for the project is the sum of the expected delay for task A plus the expected delay for task B. It is assumed the likelihood of a problem in any task is independent of the other.

Then, each expected delay for a task is equal to the probability of a problem multiplied by the consequence (delay time).

[tex]E(D)=E(D_a)+E(D_b)=p_aT_a+p_bT_b\\\\E(D)=0.5*2+0.2*3=1+0.6=1.6[/tex]

The risk consequence of both activities is 1.6 months.

Answer: C represents a function

Step-by-step explanation: It represents a function because there is exactly one range for each domain, and only one. In mathematics, an algebraic function is a function that can be defined as the root of a polynomial equation. Quite often algebraic functions are algebraic expressions using a finite number of terms, involving only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power.

Answer:

4 taps take 70 minutes

2 taps would take 140 minutes

8 taps would take 35 minutes

time = (4 / # of taps) * 70

time = (4 / 7) * 70 = 40 minutes

Step-by-step explanation:

Answer:

The proportion of semester students returning to summer school

p = 0.35

Step-by-step explanation:

Explanation:-

Given data A university planner is interested in determining the percentage of spring semester students who will attend summer school.

She takes a pilot sample of 160 spring semester students discovering that 56 will return to summer school

Given the sample size 'n' = 160

let 'x' = 56

Sample proportion of semester students returning to summer school

[tex]p = \frac{x}{n} = \frac{56}{160}[/tex]

p = 0.35

Answer:2

4 th root of 1024 is 5

Since 4 x4x4x4x4=1024

So 1+2x=5

-1 PN both sides

2x= 4

Divide by 2

X=2

Step-by-step explanation:

Answer:

69.50%

Step-by-step explanation:

Given:  

Ayrshire cows:  

E(X) =μ  = 47

SD(X) = σ = 6 Var (X) = 6^2 = 36  

Jersey cows:  

E(Y) = μ = 43

SD(Y) = σ  = 5 Var(Y) = 52 = 25

Properties mean, variance and standard deviation:  

E(X +Y) = E(X) E(Y)

V ar(X +Y) = Var(X) + Var(Y)

SD(X +Y) = √Var(X)+Var(Y)

X — Y represents the difference between Ayrshire and Jersey cows.  

E(X — Y) = E(X) — E(Y). 47 — 43 = 4

SD(X — Y) = √Var(X)+ Var(Y) =√36+ 25 = √61 = 7.8102

The z-score is the value decreased by the mean, divided by the standard deviation:  

z = x-μ /σ = 0-4/ 7.8102  = -0.51

Determine the corresponding probability using table Z in appendix F.  

P(X—Y [tex]\geq[/tex] 0) = P(Z > —0.51) = 1—P(Z < —0.51) = 1-0.3050 = 0.6950 = 69.50%  

Answer:

Galloping speed Y=179.984-0.076 weight X

Step-by-step explanation:

The required equation is

y=a+bx

where y is galloping speed and x is weight of animal.

We have to estimate intercept a and slope b.

x                  y         xy       x²

25    191.5 4787.5 625

35     184.7 6464.5 1225

50    175.8 8790 2500

75     162.2 12165 5625

500  124.9 62450 250000

1000 113.2 113200 1000000

sumx=1685

sumy=952.3

sumxy=207857

sumx²=1259975

[tex]slope=b=\frac{n(sumxy)-(sumx)(sumy)}{nsumx^{2}-(sumx)^{2} }[/tex]

[tex]slope=b=\frac{6(207857)-(1685)(952.3)}{6*(1259975)-(1685)^{2} }[/tex]

[tex]slope=b=\frac{-357483.5}{4720625}[/tex]

slope=b=-0.07573.

intercept=a=ybar-b*xbar

[tex]ybar=\frac{sumy}{n}[/tex]

[tex]ybar=\frac{952.3}{6}[/tex]

ybar=158.7167

[tex]xbar=\frac{sumx}{n}[/tex]

[tex]xbar=\frac{1685}{6}[/tex]

xbar=280.8333

intercept=a=ybar-b*xbar

intercept=a=158.7167-(-0.07573)*280.8333

intercept=a=158.7167+21.2675

intercept=a=179.9842

rounding intercept and slope to 3 decimal places

intercept=179.984.

slope=-0.076

The required equation is

Galloping speed Y=179.984-0.076 weight X

The required equation is  galloping speed Y = 179.984 - 0.076 weight X.

Given that,

Four-legged animals run with two different types of motion: trotting and galloping.

The number of strides per minute at which an animal breaks from a trot to a gallop depends on the weight of the animal.

We have to determine,

An equation that relates an animal's weight x (in pounds) and its lowest galloping speed y (in strides per minute).

According to the question,

A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation.

The required linear equation is,

y = a + bx

Where y is galloping speed and x is weight of animal.

To estimate intercept a and slope b from the table showing below.

x        y         x.y           x²

25     191.5   4787.5   625

35     184.7   6464.5  1225

50     175.8   8790     2500

75      162.2  12165     5625

500   124.9   62450  250000

1000  113.2   113200  1000000

Then,

sum of x = 1685 , sum of y = 952.3 , sum of xy = 207857 , sum of x² = 1259975.

Therefore, Slope of b is given by the formula;

[tex]b = \dfrac{(sum\ of \ xy )- (sumx).(sumy)}{n(sumx^{2})- (sumx)^{2} }\\\\b = \dfrac{6(207857)-1685(952.3)}{6(1259975)-(1685)^{2} }\\\\b = \dfrac{-357483.5}{4720625}\\\\b = -0.7573[/tex]

Then,

[tex]Y_b_a_r = \dfrac{sumx}{n} = \dfrac{952.3}{6} = 158.71\\\\X_b_a_r = \dfrac{sumy}{n} = \dfrac{1685}{6} = 280.83[/tex]

Therefore, The intercept is given by,

[tex]intercept = a = Y_b_a_r-b\times X_b_a_r\\\\intercept = a = 158.7167 - (-0.07573)\times 280.8333\\\\intercept = a = 158.7167 + 21.2675\\\\intercept = a = 179.984[/tex]

On rounding intercept and slope to 3 decimal places,

intercept = 179.984.

slope = -0.076

Hence, The required equation is  galloping speed Y = 179.984 - 0.076 weight X.

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Answer:

(0.339, 0.391)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 1295, \pi = \frac{473}{1295} = 0.365[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.365 - 1.96\sqrt{\frac{0.365*0.635}{1295}} = 0.339[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.365 + 1.96\sqrt{\frac{0.365*0.635}{1295}} = 0.391[/tex]

So the correct answer is:

(0.339, 0.391)

So, the required length of the sides that have standard fencing is 1500 ft.

The area of rectangle is the region covered by the rectangle in a two-dimensional plane. A rectangle is a type of quadrilateral, a 2d shape that has four sides and four vertices.

Let [tex]x[/tex] ft be the length of the sides that duty fencing and [tex]y[/tex] ft be the length of the sides that have standard fencing.

So, the area will be calculated by the above formula we get,

[tex]Area=xy[/tex]...(1)

Now, the cost of fencing is,

[tex]3x+2y=6000\\y=3000-\frac{3x}{2}[/tex]...(2)

Now, substituting equation (2) in equation (1) we get,

[tex]A=3000x-\frac{3x^2}{2}[/tex]

Now, differentiating the above equation we get,

[tex]\frac{dA}{dx} =3000-3x=0\\x=1000[/tex]

Substituting [tex]x=1000 ft[/tex] in equation (2) we get,

[tex]y=3000-\frac{3\times 1000}{2}\\=1500 ft[/tex]

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The question pertains to maximizing the area of a rectangular land plot within a fence budget. We derive a mathematical equation to fit the budget involving both types of fences and then maximize the equation to find the solution. It involves applications of maximization and differentiation.

The subject of this question is Mathematics, specifically problems involving equations and budgets. This problem can be solved by using the concept of maximization of areas in rectangle and budget calculations.

We know the total budget which is $6000, and costs of the two types of fencing. Let's denote the lengths of the heavy-duty fenced sides as H and the lengths of the standard fenced sides as S. The equation that describes the budget is 3H + 2S = 6000.

If we're maximizing area, we want to maximize S*H. Remember that the area of a rectangle is the product of its dimensions (length times width). Since H and S are lengths of sides of a rectangular plot, H*S will give us the area.

Replacing H from the budget equation we get H = (6000 - 2S) / 3. You can plug this into the area equation to get Area = S * (6000 - 2S) / 3. To find the maximum of this function, take its derivative and set it to zero, solve for S.

Doing this results in the required amount of standard fencing for the project, making it a solution to the question asked.

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Answer:

k= 13*3-2

k= 39-2

k=37 is the required answer

Answer:

k = 37

Step-by-step explanation:

[tex]k = 13t - 2 \\ \therefore \: k = 13 \times 3 - 2 \\ \therefore \: k = 39 - 2 \\ \huge \red{ \boxed{ \therefore \: k = 37}}[/tex]

Answer:

Yes, it is arithmetic sequence.

Step-by-step explanation:

31-26=5

36-31=5

41-36=5

46-41=5

Difference between consecutive numbers is the same, so we have arithmetic sequence.

Answer:

Yes it is an arithmetic sequence

26+5=31

31+5=36

36+5=41

41+5=46

Step-by-step explanation:

W(x) = (10x⁴ − 8) (30x + 25)^0.5

A) Take log of both sides.

ln(W) = ln[(10x⁴ − 8) (30x + 25)^0.5]

ln(W) = ln(10x⁴ − 8) + ln[(30x + 25)^0.5]

ln(W) = ln(10x⁴ − 8) + 0.5 ln(30x + 25)

Take derivative.

W' / W = 40x³ / (10x⁴ − 8) + 0.5 (30) / (30x + 25)

W' / W = 20x³ / (5x⁴ − 4) + 3 / (6x + 5)

W' = W [20x³ / (5x⁴ − 4) + 3 / (6x + 5)]

W'(x) = (10x⁴ − 8) (30x + 25)^0.5 [20x³ / (5x⁴ − 4) + 3 / (6x + 5)]

B) Evaluate at x = 0.

W'(0) = (0 − 8) (0 + 25)^0.5 [0 / (0 − 4) + 3 / (0 + 5)]

W'(0) = (-8) (5) (0 + 3/5)

W'(0) = -24

Answer:

[tex]\textsf{A)}\quad W'(x) = (10x^4 - 8) \cdot (30x + 25)^{0.5} \left(\dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5} \right)[/tex]

[tex]\textsf{B)}\quad W'(0)=-24[/tex]

Step-by-step explanation:

Given function:

[tex]W(x) = (10x^4 - 8) \cdot (30x + 25)^{0.5}[/tex]

Take the natural logarithm of both sides:

[tex]\ln \left(W(x)\right) = \ln \left( (10x^4 - 8) \cdot (30x + 25)^{0.5} \right)[/tex]

Use properties of logarithms to simplify:

[tex]\ln \left(W(x)\right) = \ln (10x^4 - 8) +\ln(30x + 25)^{0.5} \\\\\\ \ln \left(W(x)\right) = \ln (10x^4 - 8) +0.5 \ln(30x + 25)[/tex]

Differentiate both sides with respect to x:

[tex]\dfrac{d}{dx} \ln \left(W(x)\right) = \dfrac{d}{dx} \left( \ln (10x^4 - 8) + 0.5 \ln (30x + 25) \right)[/tex]

Apply the chain rule on the right-hand side:

[tex]\dfrac{W'(x)}{W(x)} = \dfrac{1}{10x^4 - 8} \cdot \dfrac{d}{dx} (10x^4 - 8) + 0.5 \cdot \dfrac{1}{30x + 25} \cdot \dfrac{d}{dx} (30x + 25) \\\\\\ \dfrac{W'(x)}{W(x)} = \dfrac{1}{10x^4 - 8} \cdot (40x^3) + 0.5 \cdot \dfrac{1}{30x + 25} \cdot (30) \\\\\\ \dfrac{W'(x)}{W(x)} = \dfrac{40x^3}{10x^4 - 8} + \dfrac{15}{30x + 25} \\\\\\ \dfrac{W'(x)}{W(x)} = \dfrac{2(20x^3)}{2(5x^4 - 4)} + \dfrac{5(3)}{5(6x + 5)} \\\\\\ \dfrac{W'(x)}{W(x)} = \dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5}[/tex]

Multiply both sides by W(x) to solve for W'(x):

[tex]W'(x) = W(x) \left(\dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5} \right)[/tex]

Substitute W(x) back in:

[tex]W'(x) = (10x^4 - 8) \cdot (30x + 25)^{0.5} \left(\dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5} \right)[/tex]

Therefore, the derivative W'(x) is:

[tex]W'(x) = (10x^4 - 8) \cdot (30x + 25)^{0.5} \left(\dfrac{20x^3}{5x^4 - 4} + \dfrac{3}{6x + 5} \right)[/tex]

[tex]\dotfill[/tex]

To find W'(0), substitute x = 0 into W'(x):

[tex]W'(0) = (10(0)^4 - 8) \cdot (30(0) + 25)^{0.5} \left(\dfrac{20(0)^3}{5(0)^4 - 4} + \dfrac{3}{6(0) + 5} \right) \\\\\\ W'(0) = (0 - 8) \cdot (0 + 25)^{0.5} \left(\dfrac{0}{0 - 4} + \dfrac{3}{0 + 5} \right) \\\\\\ W'(0) = (- 8) \cdot (25)^{0.5} \cdot \left( \dfrac{3}{5} \right) \\\\\\ W'(0) = (- 8) \cdot 5 \cdot \dfrac{3}{5} \\\\\\ W'(0)=-24[/tex]

Therefore, the value of W'(0) is -24.

Answer:

the answer is 75% hope that helps

Step-by-step explanation:

i did the test and i got it right

The required percentage of the data values falls between the values of 27 and 45 in the data set shown is 75%.

A straightforward method of expressing statistical data on a plot in which a rectangle is drawn to represent the second and third quartiles, with a vertical line inside to indicate the median value. Horizontal lines on both sides of the rectangle show the lower and upper quartiles.

Based on the box-and-whisker plot, we can see that the majority of the data values fall between the values of 32 and 45, and the median value is 36.

To find the percentage of data values that fall between 27 and 45, we can estimate by visually analyzing the plot. Since the whiskers go from 27 to 48, we can assume that all data values between 27 and 48 are included. Then, we need to estimate how much of the data falls between 32 and 45, which is the range of the box.

Since the box takes up most of the range of values between 32 and 45, we can estimate that around 75% of the data values fall within this range. Therefore, the percentage of data values that fall between 27 and 45 can be estimated as around 75%.

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Question in order:

The exact numbers of each color of Umbrella Corporation's most popular candy, W&Ws, naturally vary from bag to bag, but the company wants to make sure that the process as a whole is producing the five colors in equal proportions. Which of the following would represent the alternative hypothesis in a chi-squared goodness-of-fit test conducted to determine if the five colors of W&Ws occur in equal proportions?

a. H1: All of the proportions are different from 0.20

b. HI: p1 = P2 = p3 =P4=p5 = 0.20

c. H1: At least one proportion is not equal to 0.20

d. H1: At least two of the mean number of colors differ from one another.

e. HI: the number of candies per bag and candy color are dependent

Answer:

option C

Step-by-step explanation:

H1: At least one proportion is not equal to 0.20

Answer:

This question answer is attached in the attachment,

Step-by-step explanation:

The height of the cylinder with a volume of 384π cubic inches and a radius of 8 inches is 6 inches.

To find the height of a cylinder with a known volume and radius, we use the formula for the volume of a cylinder, which is V =  πr²h, where 'V' is the volume, 'r' is the radius, and 'h' is the height. We are given a volume of 384π cubic inches and a radius of 8 inches. Plugging these values into the formula gives us:

384 π = π (8²) h
384 π = 64π h
h = 384 π ÷ 64π
h = 6 inches

Therefore, the height of the cylinder is 6 inches.

Answer: The required probability is 0.674.

Step-by-step explanation:

Since we have given that

Number of students in the Spanish class = 38

Number of students in the French class = 27

Number of students in the German class = 16

Number of students in both spanish and French = 14

Number of students in both Spanish and German = 6

Number of students in both French and German = 5

Number of students in all three class = 2

So, it becomes:

[tex]n(S\cup F\cup G)=n(S)+n(F)+n(G)-n(S\cap F)-n(G\cap F)-n(S\cap G)+n(S\cap F\cap G)\\\\n(S\cup F\cup G)=38+27+16-14-5-6+2=58[/tex]

So, Probability that he or she is taking at least one language class is given by

[tex]\dfrac{58}{86}=0.674[/tex]

Hence, the required probability is 0.674.

Answer:

D) $896,053

Step-by-step explanation:

Answer:

Step-by-step explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ ≥ 50000

For the alternative hypothesis,

µ < 50000

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ)/(σ/√n)

Where

x = lifetime of the tyres

µ = mean lifetime

σ = standard deviation

n = number of samples

From the information given,

µ = 50000 miles

x = 45800 miles

σ = 8000

n = 29

z = (50000 - 45800)/(8000/√29) = - 2.83

Looking at the normal distribution table, the probability corresponding to the z score is 0.9977

Since alpha, 0.05 < than the p value, 0.9977, then we would accept the null hypothesis. Therefore, At a 5% level of significance, the data is not highly consistent with the claim.

Answer:

x = 10

y = 9

Step-by-step explanation:

First, you need to write two equations that represent the information given. I'm going to set the two numbers and x and y.

x - y = 1

x + y = 19

This turns it into a systems of equations. When you add the two equations, keeping everything on each side, you eliminate a variable, and solve for the other.

2x = 20

x = 10

Now, you'd plug x into either original equation.

(10) - y = 1

-y = -9

x = 10

I hope this helped!

Answer:

The numbers are 9 and 10.  Add to get a sum of 19, but subtract to get a difference of 1.  

Step-by-step explanation:

Answer:

The 98% confidence interval for the average credit card balance is

(564.04, 635.96).

Step-by-step explanation:

We have to calculate the 98% confidence interval on the average credit card balance.

The sample will consist of the n=30 customers that have credit card.

The sample has a mean of $600 and a standard deviation of $80.

As the population standard deviation is estimated from the sample standard deviation, we will use a t statistic.

The degrees of freedom are:

[tex]df=n-1=30-1=29[/tex]

The critical value for a 98% CI and 29 degrees of freedom is t=2.463 (this can be looked up in a t-table).

Then, the margin of error is:

[tex]E=t\cdot s/\sqrt{n}=2.463*80/\sqrt{30}=197.04/5.48=35.96[/tex]

Then, the upper and lower bounds of the confidence interval are:

[tex]LL=\bar X-E=600-35.96=564.04\\\\UL=\bar X+E=600+35.96=635.96[/tex]

Answer:

50 cents or 0.5

Step-by-step explanation:

Answer:

[tex] \tan \theta = - \frac{1}{3}[/tex]

Step-by-step explanation:

Since, point (-3, 1) is on the terminal side of the angle in standard position.

[tex] \therefore \: ( - 3, \: 1) = (x, \: y) \\ \therefore \:x = - 3 \: \: and \: \: y = 1 \\ \because \tan \theta = \frac{y}{x} \\ \therefore \:\tan \theta = \frac{1}{ - 3} \\ \\ \huge \red{ \boxed{\therefore \:\tan \theta = - \frac{1}{3} }}[/tex]

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable X is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, p = 0.09.

An item is defective irrespective of the others.

The random variable X follows a Binomial distribution with parameters n = 9 and p = 0.09.

The repair cost of the item is given by:

[tex]C=3X^{2}+X+2[/tex]

Compute the expected cost of repair as follows:

[tex]E(C)=E(3X^{2}+X+2)[/tex]

        [tex]=3E(X^{2})+E(X)+2[/tex]

Compute the expected value of X as follows:

[tex]E(X)=np[/tex]

         [tex]=4\times 0.09\\=0.36[/tex]

The expected value of X is 0.36.

Compute the variance of X as follows:

[tex]V(X)=np(1-p)[/tex]

         [tex]=4\times 0.09\times 0.91\\=0.3276\\[/tex]

The variance of X is 0.3276.

The variance can also be computed using the formula:

[tex]V(X)=E(Y^{2})-(E(Y))^{2}[/tex]

Then the formula of [tex]E(Y^{2})[/tex] is:

[tex]E(Y^{2})=V(X)+(E(Y))^{2}[/tex]

Compute the value of [tex]E(Y^{2})[/tex] as follows:

[tex]E(Y^{2})=V(X)+(E(Y))^{2}[/tex]

          [tex]=0.3276+(0.36)^{2}\\=0.4572[/tex]

The expected repair cost is:

[tex]E(C)=3E(X^{2})+E(X)+2[/tex]

         [tex]=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73[/tex]

Thus, the expected repair cost is $3.73.

Answer:

[tex]1100 \text{feet}^3[/tex]

Step-by-step explanation:

GIVEN: A swimming pool is circular with a [tex]20\text{ feet}[/tex] diameter. The depth is constant along east-west lines and increases linearly from [tex]1\text{ feet}[/tex] at the south end to [tex]6\text{ feet}[/tex] at the north end.

TO FIND: Find the volume of water in the pool.

SOLUTION:

Consider the image attached.

when two similar figures are attached a new cylinder is formed. volume of swimming pool is half of volume of new cylinder formed.

radius of new cylinder [tex]=\frac{\text{diameter}}{2}=\frac{20}{2}=10\text{ feet}[/tex]

height of new cylinder [tex]=6+1=7\text{ feet}[/tex]

volume of cylinder [tex]=\pi r^2h=\frac{22}{7}\times(10)^27[/tex]

                               [tex]=2200\text{ feet}^3[/tex]

Volume of swimming pool [tex]=\frac{\text{volume of cylinder}}{2}=\frac{2200}{2}[/tex]

                                           [tex]=1100\text{ feet}^3[/tex]

Hence volume of water in the pool is [tex]1100 \text{feet}^3[/tex].

Final answer:

To find the volume of water in a swimming pool with varying depths, calculate the volume by considering different height sections of the pool.The volume of water in the pool is approximately 13,287 ft³.

Explanation:

A swimming pool is circular with a 20-ft diameter. The depth is constant along east-west lines and increases linearly from 1 ft at the south end to 6 ft at the north end.

Given the pool's dimensions, you can calculate the volume of water in the pool by breaking it into sections of different heights.

The volume of water in the pool is approximately 13,287 ft³.

Answer:

[tex](0.08-0.0888) - 1.96 \sqrt{\frac{0.08(1-0.08)}{11545} +\frac{0.088(1-0.088)}{4691}}= -0.0175[/tex]  

[tex](0.08-0.0888) + 1.96 \sqrt{\frac{0.08(1-0.08)}{11545} +\frac{0.088(1-0.088)}{4691}}= 0.0015[/tex]  

And the 95% confidence interval would be given (-0.0175;0.0015).  

We are confident at 95% that the difference between the two proportions is between [tex]-0.0175 \leq p_A -p_B \leq 0.0015[/tex]

And since the confidence interval contains the 0 we have enough evidence to conclude that the population proportions are not significantly different at 5% of significance.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

[tex]p_A[/tex] represent the real population proportion for California

[tex]\hat p_A =0.08[/tex] represent the estimated proportion for California

[tex]n_A=11545[/tex] is the sample size required for California

[tex]p_B[/tex] represent the real population proportion for Oregon

[tex]\hat p_B =0.088[/tex] represent the estimated proportion for Brand B

[tex]n_B=4691[/tex] is the sample size required for Oregon

[tex]z[/tex] represent the critical value for the margin of error  

Solution to the problem

The sample proportion have the following distribution  

[tex]\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]  

The confidence interval for the difference of two proportions would be given by this formula  

[tex](\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}[/tex]  

For the 95% confidence interval the value of [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2=0.025[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.  

[tex]z_{\alpha/2}=1.96[/tex]  

And replacing into the confidence interval formula we got:  

[tex](0.08-0.0888) - 1.96 \sqrt{\frac{0.08(1-0.08)}{11545} +\frac{0.088(1-0.088)}{4691}}= -0.0175[/tex]  

[tex](0.08-0.0888) + 1.96 \sqrt{\frac{0.08(1-0.08)}{11545} +\frac{0.088(1-0.088)}{4691}}= 0.0015[/tex]  

And the 95% confidence interval would be given (-0.0175;0.0015).  

We are confident at 95% that the difference between the two proportions is between [tex]-0.0175 \leq p_A -p_B \leq 0.0015[/tex]

And since the confidence interval contains the 0 we have enough evidence to conclude that the population proportions are not significantly different at 5% of significance.