A simple random sample of 26 precipitation amounts has a standard deviation of 0.18. Find the test statistic, using a 0.01 significance level to test a claim that the standard deviation of all precipitation amounts is equal to 0.25

Answers

Answer 1

Answer:

[tex] t=(26-1) [\frac{0.18}{0.25}]^2 =12.96[/tex]

What is the critical value for the test statistic at an α = 0.01 significance level?

Since is a two tailed test the critical zone have two zones. On this case we need a quantile on the chi square distribution with 25 degrees of freedom that accumulates 0.005 of the area on the left tail and 0.995 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be [tex]\Chi^2 =10.520[/tex]    

And the right critical value would be :  [tex]\Chi^2 =46.927[/tex]

And the rejection zone would be: [tex] \chi^2 < 10.52 \cup \chi^2 >46.927[/tex]

Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

n = 26 sample size

s= 0.18

[tex]\sigma_o =0.25[/tex] the value that we want to test

[tex]p_v [/tex] represent the p value for the test

t represent the statistic

[tex]\alpha=0.01[/tex] significance level

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is equal to 0.25, so the system of hypothesis are:

H0: [tex]\sigma =0.25[/tex]

H1: [tex]\sigma \neq 0.25[/tex]

In order to check the hypothesis we need to calculate the statistic given by the following formula:

[tex] t=(n-1) [\frac{s}{\sigma_o}]^2 [/tex]

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

[tex] t=(26-1) [\frac{0.18}{0.25}]^2 =12.96[/tex]

What is the critical value for the test statistic at an α = 0.01 significance level?

Since is a two tailed test the critical zone have two zones. On this case we need a quantile on the chi square distribution with 25 degrees of freedom that accumulates 0.005 of the area on the left tail and 0.995 on the right tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.005,25)". And our critical value would be [tex]\Chi^2 =10.520[/tex]

And the right critical value would be :  [tex]\Chi^2 =46.927[/tex]

And the rejection zone would be: [tex] \chi^2 < 10.52 \cup \chi^2 >46.927[/tex]

Since our calculated value is NOT in the rejection zone we FAIL to reject the null hypothesis.

Answer 2

At the 0.01 significance level, we fail to reject the null hypothesis, meaning there is not enough evidence to refute the claim that the standard deviation is 0.25.

To test the claim that the standard deviation of all precipitation amounts is equal to 0.25 using a 0.01 significance level, we use the chi-square test for standard deviation.

The null hypothesis (H0) states that the population standard deviation (σ) is 0.25. The alternative hypothesis (Ha) states that σ is not equal to 0.25.

The test statistic for a chi-square test is calculated using the formula:

Chi-square (χ²) = (n - 1)s² / σ₀²

Where:

n is the sample size (26)s is the sample standard deviation (0.18)σ₀ is the hypothesised population standard deviation (0.25)

Substituting the values:

χ² = (26 - 1)(0.18)² / (0.25)²

Calculating this:

χ² = 25 * 0.0324 / 0.0625 = 12.96

This χ² value is compared to the critical values from the chi-square distribution table with (n-1) or 25 degrees of freedom at a 0.01 significance level. The two-tailed critical values for 25 degrees of freedom at 0.01 significance level are approximately 10.85 and 46.93.

Since 10.85 < 12.96 < 46.93, we fail to reject the null hypothesis.

Hence, at the 0.01 significance level, there is not enough evidence to reject the claim that the standard deviation of all precipitation amounts is 0.25.


Related Questions

3/5 of your class earned an A or a B last week. 1/4 of those teammates earned an A. what fraction of you class earned an A? Possible answers could be 4/9, 7/20, 2/1, 3/20 but which one is it?

Answers

Answer:

3/20

Step-by-step explanation:

3/5 earned an A or B = "3/5 of the class"1/4 of 3/5 earned an A = "3/5 of class divided by 4"

[tex]\frac{3}{5}[/tex]  ÷  4    =     [tex]\frac{3}{5}[/tex] [tex]* \frac{1}{4}[/tex]    =      [tex]\frac{3 * 1}{5 * 4}[/tex]     =     [tex]\frac{3}{20}[/tex]

Final answer:

To calculate the fraction of the class that earned an A, multiply 3/5 by 1/4, which equals 3/20. Thus, 3/20 of the class earned an A. This is the correct answer among the given options.

Explanation:

Calculating the Fraction of a Class That Earned an A

To find out what fraction of the class earned an A, we start with the fact that 3/5 of the class earned an A or a B.

We also know that 1/4 of those students earned an A.

To determine the fraction of the entire class that earned an A, we multiply these two fractions together:

Fraction of class that earned an A = (3/5) × (1/4)

This multiplication gives us:

(3 × 1) / (5 × 4) = 3/20

Therefore, 3/20 of the class earned an A.

This fraction cannot be reduced further and it is also one of the options provided, confirming that it is the correct answer.

Seventy-five (75) customers at Crummy Burger today were entered in a drawing. Three names will be drawn, and each will win a prize. Determine the number of ways that the prizes can be distributed if first prize is a $25 gift card, second prize is a free Crummy Combo Meal, and third prize is a free Jumbo Crummy Burger.

Answers

Answer:

405150

Step-by-step explanation:

3 people are to be selected from 75 people. This could be done using the combination function, [tex]\binom{75}{3}[/tex]. But thus function only gives the number of ways of choosing any 3 out of the 75 customers. For any single selection of any 3 customers, there are 3! ways of sharing the three prizes among them. Hence the total number of ways of sharing the prizes is

[tex]\binom{75}{3}\times3! = 405150[/tex]

In fact, this is the permutation function given by

[tex]{}^{75}P_3 = \dfrac{75!}{(75-3)! = 405150}[/tex]

A standard medium pizza has a diameter of 12 inches and is cut into 8 slices.
What is the area and what is the one sector area of a slice of pizza.

Please show work. I've been stuck on this for a while and want to understand it.

Answers

Answer:

The answer to your question is Circle's area = 113.04 in²  Slice' area = 14.13 in²

Step-by-step explanation:

Data

diameter = 12 in

8 slices

Area = ?

Area of a slice = ?

Process

1.- Calculate the area of the pizza

radius = 12/2

radius = 6 in

Area = πr²

-Substitution

Area = (3.14)(6)²

-Simplification

Area = 113.04 in²

2.- Calculate the area of the slice

We can calculate this area by two methods

a) Divide the area of the number of slices

       Area of the slice = 113.04 / 8

                                   = 14.13 in²

b) Using the area of a sector

-Find the angle of each slice

               360 / 8 = 45°

-Convert 45° to rad

                180° ------------------- πrad

                  45° ------------------- x

                  x = 45πrad/180

                  x = 1/4πrad

- Calculate the area of the sector

Area = 1/4(3.14)(6)²/2

Area = 113.04/8

Area = 14.13 in²                        

Answer:

Step-by-step explanation:

Assuming the pizza is circular, we would apply the formula for determining the area of a circle. It is expressed as

Area = πr²

Where

r represents the radius of the circle.

π is a constant whose value is 3.14

From the information given,

Diameter = 12 inches

Radius = 12/2 = 6 inches

Area = 3.14 × 6² = 113.04 inches²

The formula for determining the area of a sector is expressed as

Area of sector = θ/360 × πr²

Where θ represents the central angle.

The complete circle or pizza is 360°. Since it is divided into 8 equal parts, then the central angle formed by each sector is

360/8 = 45°

Therefore,

Area of sector = 45/360 × 3.14 × 6²

= 14.13 inches²

A section of an exam contains four True-False questions. A completed exam paper is selected at random, and the four answers are recorded. Assuming the outcomes to be equally likely, find the probability that all the answers are the same.

Answers

Final answer:

The probability that all the answers are the same in the section of the exam is 1/8 or 12.5%.

Explanation:

To find the probability that all the answers are the same, we need to consider the number of ways in which all the answers can be the same, divided by the total number of possible outcomes. Since there are only two possible answers (True or False) for each question, there are a total of 2^4 = 16 possible outcomes.

Out of these 16 outcomes, there are only two ways in which all the answers can be the same: either all True or all False.

Therefore, the probability that all the answers are the same is 2/16 = 1/8 = 0.125, or 12.5%.

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Find the arc length of AB. Round your answer to the nearest hundredth.

Answers

The arc length of AB = 8.37 meters

Solution:

Degree of AB (θ) = 60°

Radius of the circle = 8 m

Let us find the arc length of AB.

Arc length formula:

[tex]$\text{Arc length}=2 \pi r\left(\frac{\theta}{360^\circ}\right)[/tex]

                 [tex]$=2 \times 3.14 \times 8 \left(\frac{60^\circ}{360^\circ}\right)[/tex]

                 [tex]$=2 \times 3.14 \times 8 \left(\frac{1}{6}\right)[/tex]

Arc length = 8.37 m

Hence the arc length of AB is 8.37 meters.

A teen chewing bubble gum blows a huge bubble, the volume of which satisfies the differential equation:dV/dt=2(V)^(0.5)Where t is the time in seconds after the teen start to blow the bubble. If the bubble pops as soon as it reaches 529 cubic centimeters in volume, how many seconds does it take for the bubble to pop? Assume that the bubble had no volume when the teen first started blowing.

Answers

The ODE is separable:

[tex]\dfrac{\mathrm dV}{\mathrm dt}=2\sqrt V\implies\dfrac{\mathrm dV}{2\sqrt V}=\mathrm dt[/tex]

Integrate both sides to get

[tex]\sqrt V=t+C[/tex]

Assuming the bubble starts with zero volume, so that [tex]V(0)=0[/tex], we find

[tex]\sqrt0=0+C\implies C=0[/tex]

Then the volume of the bubble at time [tex]t[/tex] is

[tex]V(t)=t^2[/tex]

and so the time it take for the bubble to reach a volume of 629 cubic cm is

[tex]t^2=529\implies t=23[/tex]

or 23 seconds after the teen first starts blowing the bubble.

A company is considering producing some new Gameboy electronic games. Based on past records, management believes that there is a 70 percent chance that each of these will be successful and a 30 percent chance of failure. Market research may be used to revise these probabilities. In the past, the successful products were predicted to be successful based on market research 90 percent of the time. However, for products that failed, the market research predicted these would be successes 20 percent of the time. If market research is performed for a new product, what is the probability that the results indicate an unsuccessful market for the product and the product is actually unsuccessful

Answers

Final answer:

The probability that the market research predicts an unsuccessful market for the product and the product is actually unsuccessful is 24 percent. This is calculated using the concept of conditional probability and multiplying the probability of the product actually failing (30%) with the probability of market research correctly predicting a failure (80%).

Explanation:

To calculate the probability that the market research indicates an unsuccessful market for the product and the product is actually unsuccessful, we first need to understand the concept of conditional probability, given certain conditions.

Here, there is a 30 percent chance that a product will fail. The market research incorrectly predicts success for failed products 20 percent of the time. Therefore, the market research predicts failure for failed products 80 percent of the time (100% - 20%).

By multiplying these probabilities together, we can find the joint probability that a product is unsuccessful and market research also predicts it to be unsuccessful.

The result is 0.30 x 0.80 = 0.24. Therefore, there is a 24 percent chance that the market research will predict a failure and the product will also actually be a failure.

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The probability that market research indicates an unsuccessful market and the product is unsuccessful is 24%.

To calculate the probability that the results indicate an unsuccessful market for the product and that the product is unsuccessful, we need to use Bayes' Theorem. Initially, we have the probability of a product being successful, P(Success) = 0.70, and the probability of a product being a failure, P(Failure) = 0.30. Furthermore, if the product is successful, market research predicts it successful 90% of the time. If the product is a failure, market research incorrectly predicts it as a success 20% of the time, meaning it correctly predicts failure 80% of the time (100% - 20%).

We are interested in P(Market Research Failure and Actual Failure). This is the probability that the market research indicates a failure and the product indeed fails. Using the probabilities provided:

P(Market Research Successful | Success) = 0.90P(Market Research Successful | Failure) = 0.20P(Market Research Failure | Success) = 0.10P(Market Research Failure | Failure) = 0.80

Therefore, P(Market Research Failure and Actual Failure) can be calculated as:

P(Market Research Failure and Actual Failure) = P(Market Research Failure | Failure) x P(Failure) = 0.80 x 0.30 = 0.24 or 24%.

The mean and the standard deviation of the sampled population are 225.8and 17.8 respectively. What is the mean and standard deviation of the sample mean when n = 25?

Answers

Answer:

Mean and standard deviation of the sample mean are 225.8 and 3.56 respectively.

Step-by-step explanation:

The mean (μₓ) and standard deviation of the sample mean (σₓ) are related to the mean (μ) and standard deviation of the population (σ) through the following relationship

μₓ = μ = 225.8

σₓ = σ/√n = 17.8/√25 = 17.8/5 = 3.56

Executives at a large multinational company are being accused of spending too much time on email and not enough time with their employees whom they supervise. In order to see if this is true, the VP of Human Resources decides to do a comparison of the number of minutes per day that the top 100 executives spend on e-mail compared to the total minutes spent by everyone in the company. The average minutes spent per day by each employee was 96 minutes. The average for just the top 100 executives was 91 minutes with a standard deviation of 25 minutes. On average, are the Executives spending more time on e-mail than all the employees? If you wanted to use the the 1% level of significance to test your hypothesis, what would be the correct critical value?

Answers

Answer:

We conclude that there is not enough evidence to support the claim that company is spending too much time on email.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 96 minutes

Sample mean, [tex]\bar{x}[/tex] = 91 minutes

Sample size, n = 100

Alpha, α = 0.05

Sample standard deviation, s = 25 minutes

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu \leq 96\text{ minutes}\\H_A: \mu > 96\text{ minutes}[/tex]

We use one-tailed t test to perform this hypothesis.

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{91 - 96}{\frac{25}{\sqrt{100}} } = -2[/tex]

Now, [tex]t_{critical} \text{ at 0.05 level of significance, 99 degree of freedom } = 1.6603[/tex]

Since,                        

The calculated test statistic is less than the critical value, we fail to reject the null hypothesis and accept it.

Thus, we conclude that there is not enough evidence to support the claim that company is spending too much time on email.

A principal of $4700 was invested at 4.75% interest, compounded annually.
Lett be the number of years since the start of the investment. Let y be the value of the investment, in dollars.
Write an exponential function showing the relationship between y and t.

Answers

Answer:

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

y = P(1 + r/n)^nt

Where

y = the value of the investment at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount invested

From the information given,

P = $4700

r = 4.75% = 4.75/100 = 0.0475

n = 1 because it was compounded once in a year.

Therefore, the exponential function showing the relationship between y and t is

y = 4700(1 + 0.0475/1)^1 × t

y = 4700(1.0475)^t

A city park employee collected 1,850 cents in nickels, dimes, and quarters at the bottom of a wishing well. There were 30 nickels, and a combined total of 110 dimes and quarters. How many dimes and quarters were at the bottom of the wishing well?
dimes ??
quarters ???

can someone tell me how to solve this problem, ive been getting it wrong. I think I've missed a step maybe.

Answers

Answer: there are 70 dimes and 40 quarters.

Step-by-step explanation:

A dime is 10 cents

A nickel is 5 cents

A quarter is 25 cents

Let x represent the number of dimes at the bottom.

Let y represent the number of quarters at the bottom.

The city park employee collected 1,850 cents in nickels, dimes, and quarters at the bottom of a wishing well. If there were 30 nickels, it means that

10x + 25y + 30 × 5 = 1850

10x + 25y = 1850 - 150

10x + 25y = 1700- - - - - - - - -1

There is a combined total of 110 dimes and quarters. It means that

x + y = 110

Substituting x = 110 - y into equation 1, it becomes

10(110 - y) + 25y = 1700

1100 - 10y + 25y = 1700

1100 + 15y = 1700

15y = 1700 - 1100

15y = 600

y = 600/15

y = 40

x = 110 - y = 110 - 40

x = 70

Given the system of equations
(D+5)x+D2y=cost
D2x+(D+9)y=t2+3t
(a) Write the system as a differentialequation for x , in operator form, by filling thepolynomial operator in D (on the left-hand-sideof the equation) and the corresponding right-hand-side of theequation:
x=
(Write your expression for the right-hand-side as a function oft only---that is, with no derivativeoperators.)

(b) Note that we could equally well solve fory. Do this here, obtaining an equivalent differentialequation for y.
y=
(Write your expression for the right-hand-side as a function oft only---that is, with no derivativeoperators.)

Answers

The differential equations for x and y are:

x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))

y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))

Here's how to solve the problem:

(a) Write the system as a differential equation for x:

Eliminate y: Substitute the second equation into the first equation to eliminate y.

(D+5)x + D2(t2 + 3t) = cost

Rearrange and define operator: Define P(D) = (D+5) + D^2 (D+9).

P(D) x = cost - D^2 (t2 + 3t)

Write as differential equation:

x = (cost - D^2 (t2 + 3t)) / P(D)

(b) Write the system as a differential equation for y:

Eliminate x: Substitute the first equation into the second equation to eliminate x.

D2((cost - D2y) / (D+5)) + (D+9)y = t2 + 3t

Rearrange and define operator: Define Q(D) = D2 / (D+5) + (D+9).

Q(D) y = t2 + 3t - D2 cost

Write as differential equation:

y = (t2 + 3t - D2 cost) / Q(D)

Therefore, the differential equations for x and y are:

x = (cost - D^2 (t2 + 3t)) / ((D+5) + D^2 (D+9))

y = (t2 + 3t - D2 cost) / (D2 / (D+5) + (D+9))

Final Answer:

(a)[tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)\)[/tex]

(b)[tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\)[/tex]   by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations .

Explanation:

The given system of equations can be transformed into differential equations using operator (D), representing differentiation with respect to a variable, often time.

(a) To derive the differential equation for (x), we isolate (x) in terms of the differential operator \(D\). Rearranging the first equation gives us [tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\).[/tex]Substituting the expression for (y) from the second equation into this yields[tex]\(x = \frac{\cos(t)}{D+5} - \frac{D^2}{D+5}\left(\frac{t^2 + 3t}{D+9}\right)\).[/tex] Simplifying further leads to [tex]\(x = \frac{1}{D^3 + 14D + 5}\cdot \left( \cos(t) \right)[/tex]\).

(b) Similarly, isolating \(y\) from the given equations leads to [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}x\).[/tex]Substituting the expression for (x) from the first equation into this yields [tex]\(y = \frac{t^2 + 3t}{D+9} - \frac{D+5}{D+9}\left(\frac{\cos(t)}{D+5} - \frac{D^2y}{D+5}\right)\).[/tex]Further simplification results in [tex]\(y = \frac{1}{D^3 + 14D + 5}\cdot \left( t^2 + 3t \right)\).[/tex]

Therefore, by expressing (x) and (y) in terms of differential operators and given functions of (t), we derive their differential equations as specified in the final answer.

Football Strategies A particular football team is known to run 30% of its plays to the left and 70% to the right. A linebacker on an opposing team notes that the right guard shifts his stance most of the time (80%) when plays go to the right and that he uses a balanced stance the remainder of the time. When plays go to the left, the guard takes a balanced stance 90% of the time and the shift stance the remaining 10%. On a particular play, the linebacker notes that the guard takes a balanced stance.

a. What is the probability that the play will go to the left?

b. What is the probability that the play will go to the right?

c. If you were the linebacker, which direction would you prepare to defend if you saw the balanced stance?

Answers

Answer:

a) 0.659

b) 0.341

c) It'll be smarter to prepare to defend the left hand side as the probability of play going through that side when the right guard takes a balanced stance is almost double of the probability that the play would go through the right when the right guard takes a balanced stance.

Step-by-step explanation:

Considering the left first

Play goes through the left 30% of the time

- The right guard takes a balanced stance 90% of the time when play goes through the left, that is, 0.9 × 0.3 = 27% overall.

- He takes a shift stance 10% of the time that play goes through the left = 0.1 × 0.3 = 3% overall.

Play goes through the right, 70% of the time.

- He takes a balanced stance 20% of the time that play goes through the right, that is, 0.7 × 0.2 = 14% overall

- He takes a shift stance 80% of the time that play goes through the right, that is, 0.7 × 0.8 = 56% overall.

Then the right guard takes a balanced stance 27% + 14% of all the time = 41% overall.

a) Probability that play goes through the left with the right guard in a balanced stance = 27/41 = 0.659

b) Probability that play goes through the right with the right guard in a balanced stance = 14/41 = 0.341

c) It'll be smarter to prepare to defend the left hand side as the probability of play going through that side when the right guard takes a balanced stance is almost double of the probability that the play would go through the right when the right guard takes a balanced stance.

Final answer:

Using conditional probability, we can find the probabilities of the play going left or right. The probability of the play going left is 0.3, and the probability of the play going right is 0.7. If the linebacker sees a balanced stance, the probability of the play going left is approximately 0.658, so the linebacker should prepare to defend against a play going left.

Explanation:

To find the probabilities of the play going left or right, we need to use conditional probability. Let's denote L as the event of the play going left, and R as the event of the play going right. We are given P(L) = 0.3 and P(R) = 0.7. We also have the following conditional probabilities: P(B|L) = 0.9 (balanced stance when play goes left) P(S|L) = 0.1 (shift stance when play goes left) P(B|R) = 0.2 (balanced stance when play goes right) P(S|R) = 0.8 (shift stance when play goes right)

a. To find the probability that the play will go left, we can use the Law of Total Probability. P(L) = P(L∩B) + P(L∩S) = P(L)P(B|L) + P(L)P(S|L) = 0.3 X 0.9 + 0.3 * 0.1 = 0.27 + 0.03 = 0.3.

b. Similarly, to find the probability that the play will go right, P(R) = P(R∩B) + P(R∩S) = P(R)P(B|R) + P(R)P(S|R) = 0.7 X 0.2 + 0.7 X 0.8 = 0.14 + 0.56 = 0.7.

c. If the linebacker sees a balanced stance, it means that P(B) = P(B|L)P(L) + P(B|R)P(R) = 0.9 X 0.3 + 0.2 X 0.7 = 0.27 + 0.14 = 0.41. Now, we can compare the conditional probability of the play going left given a balanced stance, P(L|B) = P(B|L)P(L)/P(B) = (0.9 X 0.3)/0.41 = 0.27/0.41 ≈ 0.658. Therefore, the linebacker should prepare to defend against a play going to the left.

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The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. When traps are checked periodically, the mean number of moths per trap is only 1.2, but some traps have several moths. The distribution of moth counts in traps is strongly right skewed, with standard deviation 1.4. A random sample of 60 traps has x = 1 and s = 2.4.

Let X = the number of moths in a randomly selectd trap

(a) For the population distribution, what is the ...

...mean? =
...standard deviation? =

(b) For the distribution of the sample data, what is the ...

...mean? =
...standard deviation? =

(c) What shape does the distribution of the sample data probably have?

Exactly NormalApproximately Normal Right skewedLeft skewed



(d) For the sampling distribution of the sample mean with n = 60, what is the ...

...mean? =
...standard deviation? = (Use 3 decimal places)

(e) What is the shape of the sampling distribution of the sample mean?

Left skewedApproximately Normal Right skewedExactly Normal


(f) If instead of a sample size of 60, suppose the sample size were 10 instead. What is the shape of the sampling distribution of the sample mean for samples of size 10?

Approximately normalSomewhat left skewed Exactly NormalSomewhat right skewed


(g) Can we use the Z table to calculate the probability a randomly selected sample of 10 traps has a sample mean less than 1?

No, because the sampling distribution of the sample mean is somewhat right skewedYes, by the Central Limit Theorem, we know the sampling distribution of the sample mean is normal

Answers

Answer:

Step-by-step explanation:

Hello!

X: number of gypsy moths in a randomly selected trap.

This variable is strongly right-skewed. with a standard deviation of 1.4 moths/trap.

The mean number is 1.2 moths/trap, but several have more.

a.

The population is the number of moths found in traps places by the agriculture departments.

The population mean μ= 1.2 moths per trap

The population standard deviation δ= 1.4 moths per trap

b.

There was a random sample of 60 traps,

The sample mean obtained is X[bar]= 1

And the sample standard deviation is S= 2.4

c.

As the text says, this variable is strongly right-skewed, if it is so, then you would expect that the data obtained from the population will also be right-skewed.

d. and e.

Because you have a sample size of 60, you can apply the Central Limit Theorem and approximate the distribution of the sampling mean to normal:

X[bar]≈N(μ;σ²/n)

The mean of the distribution is μ= 1.2

And the standard deviation is σ/√n= 1.4/50= 0.028

f. and g.

Normally the distribution of the sample mean has the same shape of the distribution of the original study variable. If the sample size is large enough, as a rule, a sample of size greater than or equal to 30 is considered sufficient you can apply the theorem and approximate the distribution of the sample mean to normal.

You have a sample size of n=10 so it is most likely that the sample mean will have a right-skewed distribution as the study variable. The sample size is too small to use the Central Limit Theorem, that is why you cannot use the Z table to calculate the asked probability.

I hope it helps!

find the horizontal asymptote of the graph of y= -2x^6+5x+8/8x^6+6x+5

a: y= -1/4

b: y=0

c= y=1

D= no horizontal asymptote

NEED HELP ASAP!!!!

Answers

Answer:

a) y = -¼

Step-by-step explanation:

(-2x^6+5x+8)/(8x^6+6x+5)

Just divide the leading coefficients

-2/8 = -¼

Horizontal asymptote:

y = -¼

Answer:a) y = -¼

Step-by-step explanation:

(-2x^6+5x+8)/(8x^6+6x+5)

Just divide the leading coefficients

what is the area of a triangle expressed as a monomial

Answers

Answer:

[tex] 28x^2 [/tex]

Step-by-step explanation:

[tex]area \:o f \: \triangle = \frac{1}{2} \times 8x \times 7x \\ = 4x \times 7x \\ = 28 {x}^{2} [/tex]

Triangle area =1/2 x high x base
=1/2 X 7x X 8x= 1/2 X 56x
28x

A contractor is required by a county planning department to submit one, two, three, four, five, or six forms (depending on the nature of the project) in applying for a building permit. Let Y = the number of forms required of the next applicant. The probability that y forms are required is known to be proportional to y—that is, p(y) = ky for y = 1, , 6. (Enter your answers as fractions.) (a) What is the value of k? [Hint: 6 y = 1 p(y) = 1] k = 1/21 (b) What is the probability that at most four forms are required? 10/21 (c) What is the probability that between two and five forms (inclusive) are required?

Answers

Answer:

a)

[tex]k = \dfrac{1}{21}[/tex]

b) 0.476

c) 0.667    

Step-by-step explanation:

We are given the following in the question:

Y = the number of forms required of the next applicant.

Y: 1, 2, 3, 4, 5, 6

The probability is given by:

[tex]P(y) = ky[/tex]

a) Property of discrete probability distribution:

[tex]\displaystyle\sum P(y_i) = 1\\\\\Rightarrow k(1+2+3+4+5+6) = 1\\\\\Rightarrow k(21) = 1\\\\\Rightarrow k = \dfrac{1}{21}[/tex]

b) at most four forms are required

[tex]P(y \leq 4) = \displaystyle\sum^{y=4}_{y=1}P(y_i)\\\\P(y \leq 4) = \dfrac{1}{21}(1+2+3+4) = \dfrac{10}{21} = 0.476[/tex]

c) probability that between two and five forms (inclusive) are required

[tex]P(2\leq y \leq 5) = \displaystyle\sum^{y=5}_{y=2}P(y_i)\\\\P(2\leq y \leq 5) = \dfrac{1}{21}(2+3+4+5) = \dfrac{14}{21} = 0.667[/tex]

Final answer:

The constant of proportionality (k) equals 1/21. The probability that at most four forms are required is 10/21. The probability that between two and five forms (inclusive) are required is 14/21 or 2/3 when simplified.

Explanation:

The student has provided information regarding a probability distribution which indicates that the probability (p(y)) is proportional to the number of forms required (y), with y ranging from 1 to 6. Since we know probabilities must sum to 1, this allows us to find the constant of proportionality (k).

Part (a): Finding the Value of k

We can use the formula sum of all probabilities equals 1: 1 = k * (1 + 2 + 3 + 4 + 5 + 6). The sum of the numbers from 1 to 6 is 21, therefore 1 = 21k. Solving for k gives us k = 1/21.

Part (b): Probability of At Most Four Forms Required

To find the probability of at most four forms being required, we need to sum the probabilities for 1, 2, 3 and 4 forms: P(x ≤ 4) = k + 2k + 3k + 4k. Substituting the value of k, we get P(x ≤ 4) = (1/21)(1 + 2 + 3 + 4) = 10/21.

Part (c): Probability Between Two and Five Forms Required (Inclusive)

We calculate this by adding the probabilities for 2, 3, 4, and 5 forms: P(2 ≤ x ≤ 5) = 2k + 3k + 4k + 5k. With k = 1/21, this probability is P(2 ≤ x ≤ 5) = (1/21)(2 + 3 + 4 + 5) = 14/21 or 2/3 when simplified.

Suppose that combined verbal and math SAT scores follow a normal distribution with a mean 896 and standard deviation 174. Suppose further that Peter finds out he scored in the top 2.5 percentile (97.5% of students scored below him). Determine how high Peter's score must have been.

Answers

Let x* denote Peter's score. Then

P(X > x*) = 0.025

P((X - 896)/174 > (x* - 896)/174) = 0.025

P(Z > z*) = 1 - P(Z < z*) = 0.025

P(Z < z*) = 0.975

where Z follows the standard normal distribution (mean 0 and std dev 1).

Using the inverse CDF, we find

P(Z < z*) = 0.975  ==>  z* = 1.96

Then solve for x*:

(x* - 896)/174 = 1.96  ==>  x* = 1237.04

so Peter's score is roughly 1237.

In a recent survey, 10 percent of the participants rated Pepsi as being "concerned with my health." PepsiCo's response included a new "Smart Spot" symbol on its products that meet certain nutrition criteria, to help consumers who seek more healthful eating options. At α = .05, would a follow-up survey showing that 18 of 100 persons now rate Pepsi as being "concerned with my health" prove that the percentage has increased?

Answers

Answer:

[tex]z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67[/tex]  

[tex]p_v =P(z>2.67)=0.0038[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X=18 represent the adults who rate Pepsi as being "concerned with my health"

[tex]\hat p=\frac{18}{100}=0.18[/tex] estimated proportion of adults who rate Pepsi as being "concerned with my health"

[tex]p_o=0.10[/tex] is the value that we want to test

[tex]\alpha=0.05[/tex] represent the significance level

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.1.:  

Null hypothesis:[tex]p \leq 0.1[/tex]  

Alternative hypothesis:[tex]p > 0.1[/tex]  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.18 -0.1}{\sqrt{\frac{0.1(1-0.1)}{100}}}=2.67[/tex]  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>2.67)=0.0038[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who rate Pepsi as being "concerned with my health" i significantly higher than 0.1

The student is asked to perform a hypothesis test at the  alpha = 0.05 level to determine if the percentage of participants rating Pepsi as 'concerned with my health' has statistically increased from 10% to 18% in a follow-up survey.

The question involves conducting a hypothesis test to determine if the percentage of participants rating Pepsi as being "concerned with my health" has increased after a new initiative. With an initial percentage of 10%, a follow-up survey showed that 18 out of 100 (or 18%) now rate Pepsi in this manner. To determine if this is a statistically significant increase, at the alpha = 0.05 significance level, a hypothesis test can be performed using the binomial or normal approximation to the binomial distribution.

Null hypothesis (H0): p = 0.10 (The percentage of those who believe Pepsi is concerned with their health has not changed)
Alternative hypothesis (Ha): p > 0.10 (The percentage has increased)

If the calculated p-value is less than the significance level (0.05), we can reject the null hypothesis in favor of the alternative hypothesis, thus concluding that the percentage has indeed increased.

The materials and equipment you will have include 100% alcohol, test tubes (large diameter test tubes), fermentation tubes (smaller diameter tubes closed off at one end), yeast, and sugar. The yeast and sugar are mixed to start the fermentation process. The fermentation tube can be used to collect gases. State a testable hypothesis for your speculation from item 8 above, using the above equipment.

Answers

Answer:

A hypothesis is a proposed statement explaining the relationship between two or more variable which can be verifiable through experiment. In this case a testable hypothesis would be.

When yeast and sugar are mixed together during formation, it will cause the sugar to undergo a high rate of glycolysis and CO[tex]^{2}[/tex] will be present in the fermentation tube.

f(x) = x3 − 9x2 − 21x + 8
(a) Find the interval on which f is increasing.
Find the interval on which f is decreasing
(b) Find the local minimum and maximum values of f.
(c) Find the inflection point.
(d)Find the interval on which f is concave up.
(e)Find the interval on which f is concave down.

Answers

Final answer:

The function f(x) = x3 - 9x2 - 21x + 8 increases in the intervals (-∞, -1) and (7, ∞), and decreases in the interval (-1, 7). It achieves a local minimum at -190 and a local maximum at -23. The inflection point is at x = 3, with the function being concave up for x < 3 and concave down for x > 3.

Explanation:

Given the function f(x) = x3 - 9x2 - 21x + 8, we need to find intervals where the function is increasing or decreasing, and determine its local minima/maxima, inflection points, and where it is concave up or down.

To find the intervals, we need to first take the derivative of f(x): f'(x) = 3x2 - 18x - 21 which becomes 3(x - 7)(x + 1). The critical points are then 7 and -1.

The function increases in the interval (-∞, -1) and decreases in the interval (-1, 7). The function again increases in the interval (7, ∞).

Local minima and maxima can be found by evaluating the function at the critical points. f(-1) = -23 and f(7) = -190, so the local maximum value of f is -23 and local minimum is -190.

To find the inflection point, we compute the second derivative f''(x) = 6x - 18, which simplifies to 6(x - 3), giving us an inflection point at x = 3.

The function is concave up for x < 3 and concave down for x > 3.

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a) f(x) is increasing for x in (- ∞, - 1) cup(7, infty)

f(x) is decreasing for x \in (- 1, 7)

b)Local minimum=-237 Local maximum=19

c)Point of inflection: (x, y) = (3, - 109)

d)Interval for concave upward curve: x \in (3, ∞)

e)Interval for concave upward curve: x \in (- ∞, 3)

The given function is [tex]f(x) = x ^ 3 - 9x ^ 2 - 21x + 8[/tex]

[tex]f' * (x) = 3x ^ 2 - 18x - 21[/tex]

f^ prime prime (x) = 6x - 18

a)f(x) is increasing for

f' * (x) > 0

[tex]3x ^ 2 - 18x - 21 > 0[/tex]

[tex]x ^ 2 - 6x - 7 > 0[/tex]

[tex]x ^ 2 - 7x + x - 7 > 0[/tex]

x(x - 7) + 1(x - 7) > 0

(x - 7)(x + 1) > 0

x > 7 and x > - 1or x < 7 and x < - 1

f(x) is decreasing for

f' * (x) < 0

3x ^ 2 - 18x - 21 < 0

x ^ 2 - 6x - 7 < 0

x ^ 2 - 7x + x - 7 < 0

x(x - 7) + 1(x - 7) < 0

(x - 7)(x + 1) < 0

x > 7 and x < - 1 or x < 7 and x > - 1

x=(-1,7)

b)for local extrema

f' * (x) = 0

3x ^ 2 - 18x - 21 = 0

x ^ 2 - 6x - 7 < 0

x ^ 2 - 7x + x - 7 = 0

x(x - 7) + 1(x - 7) = 0

(x - 7)(x + 1) = 0

x = 7 and x = - 1

At x=-1:

f(- 1) = (- 1) ^ 3 - 9 * (- 1) ^ 2 - 21(- 1) + 8 = 19

At x=7:

f(7) = (7) ^ 3 - 9 * (7) ^ 2 - 21(7) + 8 = - 237

Local minimum=-237

Local maximum=19

c) For point of inflection:

f^ prime prime (x) = 0

Rightarrow 6x - 18 = 0

Rightarrow x = 3

y = f(3) = (3) ^ 3 - 9 * (3) ^ 2 - 21(3) + 8 = - 109

(x, y) = (3, - 109)

d)Interval for concave upward curve:

f^ prime prime (x) > 0

6x - 18 > 0

x \in (3, ∞)

e)Interval for concave upward curve:

f^ prime prime (x) < 0

6x - 18 < 0

X€(−0,3)

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Petroleum pollution in oceans stimulates the growth of certain bacteria. An assessment of this growth has been madew by counting the bacteria in each of 5 randomly chosen specimens of ocean water (of a fixed size). The 5 counts obtained were as follows.

41, 62, 45, 48, 69

Find the deviation of this sample of numbers. Round your answers to at least two decimal places.

Answers

Final answer:

The standard deviation of the sample of numbers is approximately 10.68, after calculating the mean, finding the deviations, squaring them, averaging the squared deviations to get the variance, and then taking the square root of the variance.

Explanation:

The student is asking how to calculate the standard deviation of a sample of numbers. The sample given consists of the counts of bacteria found in ocean water, which are 41, 62, 45, 48, 69. To calculate the standard deviation, follow these steps:

Calculate the mean (average) of the sample.Subtract the mean from each number in the sample to find their deviations.Square each deviation to get the squared deviations.Calculate the average of the squared deviations (this is the variance).Take the square root of the variance to get the standard deviation.

Performing the calculations, you get:

Mean (average): (41 + 62 + 45 + 48 + 69) / 5 = 265 / 5 = 53Deviation for each number: -12, 9, -8, -5, 16Squared deviations: 144, 81, 64, 25, 256Variance: (144 + 81 + 64 + 25 + 256) / 5 = 570 / 5 = 114Standard deviation: √114 ≈ 10.68 (rounded to two decimal places)

Find the slope of the line passing through each pair of points or state that the slope is undefined. Assume that all variables represent positive real numbers. Then indicate whether the line through the points rises, falls, is horizontal, or is vertical. (-a, 0) and (0, -b)

Answers

Answer:

-b/a and line is falling

Step-by-step explanation:

Let (-a,0) and (0,-b) be point 1 and 2, respectively. The slope passing through point (-a,0) and (0,-b) can be calculated using the following equation

[tex]m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-b - 0}{0 - (-a)} = -b/a[/tex]

As b and a are positive real number, then the slope m = -b/a is a negative real number. The line passing through these 2 points is falling.

The slope of the line passing through points (-a, 0) and (0, -b) is calculated using the formula for slope, resulting in a slope of -b/a, which indicates that the line falls as it moves from left to right.

To calculate the slope (m), we use the formula m = (y2 - y1) / (x2 - x1), with (x1, y1) = (-a, 0) and (x2, y2) = (0, -b). Substituting the values, we get m = (-b - 0) / (0 - (-a)) = -b/a. Since both a and b are positive real numbers, the sign of the slope depends on the negative numerator, indicating the line falls as it goes from left to right. If a and/or b were zero, revealing a vertical or horizontal line, a different methodology would apply, such as indicating an undefined slope for a vertical line or a zero slope for a horizontal line.

Jane and Nancy were both awarded National Merit Scholarships (for academic excellence), and both love mathematics. Jane breezed through Dr. Morgan's calculus class. Thus, Nancy should do well in that class, too."

This argument is:

a. Deductive and Valid
b. Deductive and Invalid
c. Inductive and Strong
d. Inductive and Weak

Answers

Answer:

c. Inductive and Strong

Step-by-step explanation:

In inductive reasoning, provided data is analyzed in order to reach a conclusion. In this case, the argument provides data regarding Jane and Nancy's awards and their love for mathematics and then draws a conclusion regarding Nancy's performance in a particular class, this is an example of inductive reasoning.

As for the strength of the argument, it is plausible to infer that Jane and Nancy have similar mathematics skills since they both love calculus and excel academically. Therefore, if Jane does well in the calculus class, it is a strong argument to say that Nancy does as well.

The answer is :

c. Inductive and Strong

Final answer:

The argument is Deductive and Invalid as it does not guarantee Nancy's success in the calculus class based on Jane's performance.

Explanation:

The argument presented is Deductive and Invalid. This is because the conclusion that Nancy will do well in the calculus class based on Jane's performance does not necessarily follow logically. Just because Jane excelled does not guarantee the same result for Nancy.

In deductive reasoning, the conclusion must follow necessarily from the premises, which is not the case here. An example of a valid deductive argument would be 'All humans are mortal, Socrates is a human, therefore Socrates is mortal.'

Because the argument in question does not provide a guarantee of Nancy's success based on Jane's performance, it is categorized as Invalid in the realm of deductive reasoning.

A population has mean 187 and standard deviation 32. If a random sample of 64 observations is selected at random from this population, what is the probability that the sample average will be less than 182

Answers

Answer:

11.51% probability that the sample average will be less than 182

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

In this problem, we have that:

[tex]\mu = 187, \sigma = 32, n = 64, s = \frac{32}{\sqrt{64}} = 4[/tex]

What is the probability that the sample average will be less than 182

This is the pvalue of Z when X = 182. So

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

By the Central Limit Theorem

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

[tex]Z = \frac{182 - 187}{4}[/tex]

[tex]Z = -1.2[/tex]

[tex]Z = -1.2[/tex] has a pvalue of 0.1151

11.51% probability that the sample average will be less than 182

Zenich Corp manufactures laptops. The waiting time is 72 minutes before the start of production and the manufacturing lead time is 160 minutes per laptop. What is its manufacturing cycle​ efficiency? (Round to the nearest whole​ percent.)

Answers

Answer:

The Manufacturing cycle​ efficiency = 69 %

Step-by-step explanation:

The waiting time is  = 72 minute

The manufacturing lead time is ( [tex]T_{lead}[/tex] )= 160 minutes

Total cycle time ( [tex]T_{c}[/tex] ) = 160 + 72 = 232 minutes

Manufacturing cycle​ efficiency =  [tex]\frac{T_{lead} }{T_{c} }[/tex]

⇒ Manufacturing cycle​ efficiency = [tex]\frac{160}{232}[/tex]

⇒ Manufacturing cycle​ efficiency = 69 %

Therefore, the Manufacturing cycle​ efficiency = 69 %

Find the function r that satisfies the given condition. r'(t) = (e^t, sin t, sec^2 t): r(0) = (2, 2, 2) r(t) = ()

Answers

Answer:

r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2)

Step-by-step explanation:

A primitive of e^t is e^t+c, since r(0) has 2 in its first cooridnate, then

e^0+c = 2

1+c = 2

c = 1

Thus, the first coordinate of r(t) is e^t + 1.

A primitive of sin(t) is -cos(t) + c (remember that the derivate of cos(t) is -sin(t)). SInce r(0) in its second coordinate is 2, then

-cos(0)+c = 2

-1+c = 2

c = 3

Therefore, in the second coordinate r(t) is equal to -cos(t)+3.

Now, lets see the last coordinate.

A primitive of sec²(t) is tan(t)+c (you can check this by derivating tan(t) = sin(t)/cos(t) using the divition rule and the property that cos²(t)+sin²(t) = 1 for all t). Since in its third coordinate r(0) is also 2, then we have that

2 = tan(0)+c = sin(0)/cos(0) + c = 0/1 + c = 0

Thus, c = 2

As a consecuence, the third coordinate of r(t) is tan(t) + 2.

As a result, r(t) = (e^t +1, -cos(t) + 3, tan(t) + 2).

Final answer:

To find the function r(t) that satisfies the given condition, integrate each component of r'(t) and solve for the constants using the initial condition.

Explanation:

To find the function r(t) that satisfies the condition r'(t) = (e^t, sin t, sec^2 t) and r(0) = (2, 2, 2), we need to integrate each component of r'(t) with respect to t.

The integral of e^t with respect to t is e^t + C_1, where C_1 is a constant. The integral of sin t with respect to t is -cos t + C_2, where C_2 is a constant.The integral of sec^2 t with respect to t is tan t + C_3, where C_3 is a constant.

Combining these integrals, we get r(t) = (e^t + C_1, -cos t + C_2, tan t + C_3). Substituting the initial condition r(0) = (2, 2, 2), we can solve for the constants and obtain the function r(t) = (e^t + 2, -cos t + 2, tan t + 2).

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The histogram to the right represents the weights​ (in pounds) of members of a certain​ high-school math team. How many team members are included in the​ histogram?

Answers

Answer:

Total member = a+b+c+d+e+f+g=5+4+2+0+5+0+4=20 member

Step-by-step explanation:

Let suppose the attached histogram

a) Total members whose weight 110 = 5

b) Total members whose weight 120 = 4

c) Total members whose weight 130 = 2

d) Total members whose weight 140 = 0

e) Total members whose weight 150 = 5

f) Total members whose weight 160 = 0

g) Total members whose weight 170 = 4

Total member = a+b+c+d+e+f+g=5+4+2+0+5+0+4=20 member

A histogram is a statistical graphical representation of data with the help of bars with different heights.

find the measure of exterior angle tuv below .​

Answers

Given:

m∠T = 61°

m∠S = 5x - 15

m ext∠TUV = 8x - 8

To find:

The measure of exterior angle TUV

Solution:

STU is a triangle.

By exterior angle of triangle property:

The measure of an exterior angle is equal to the sum of the opposite interior angles.

m ext∠TUV = m∠T + m∠S

8x - 8 = 61 + 5x - 15

8x - 8 = 46 + 5x

Add 8 on both sides.

8x - 8 + 8 = 46 + 5x + 8

8x = 54 + 5x

Subtract 5x from both sides.

8x - 5x = 54 + 5x - 5x

3x = 54

Divide by 3 on both sides.

[tex]$\frac{3x}{3} =\frac{54}{3}[/tex]

x = 18

Substitute x = 18 in m ext∠TUV.

m ext∠TUV = 8(18) - 8

                   = 144 - 8

                   = 136

Therefore measure of exterior angle ∠TUV is 136°.

Let X, the number of flaws on the surface of a randomly selected boiler of a certain type, have a Poisson distribution with parameter μ = 5. Use the cumulative Poisson probabilities from the Appendix Tables to compute the following probabilities. (Round your answers to three decimal places.)

(a) P(X ≤ 8)
(b) P(X = 8)
(c) P(9 ≤ X)
(d) P(5 ≤ X ≤ 8)
(e) P(5 < X < 8)

Answers

Answer:

[tex]a. \ P(X\leq 8)=0.932\\b. \ P(X=8)=0.065\\c. \ P(9\leq X)=0.068\\d. \ P(5\leq X \leq 8)=0.491\\[/tex]

e. P(5<X<8)=0.251

Step-by-step explanation:

Given that boiler follows a Poisson distribution,$\sim$

[tex]X $\sim$Poi(\mu=5)[/tex]

Poisson Distribution formula is given by the expression:-

[tex]p(x,\mu)=\frac{e^{-\mu}\mu^x}{x!}[/tex]

Our probabilities will be calculated as below:

a.

[tex]P(X=x)=\frac{5^xe^{-5}}{x!}\\P(X\leq 8)=P(X=0)+P(X=1)+...+P(X=8)\\=\frac{5^0e^{-5}}{1!}+...+\frac{5^8e^{-5}}{8!}\\=0.006738+...+0.065278\\=0.961[/tex]

b.

[tex]P(X=8)=\frac{5^8e^{-5}}{8!}\\=0.065[/tex]

c.

[tex]P(9\leq X)=1-P(X=0)-P(X=1)-...-P(X=8)\\=0.068[/tex]

d.

[tex]P(5\leq X\leq 8)=P(X=5)+P(X=6)+P(X=7)+P(X=8)\\=0.491[/tex]

e.P(5<X<8)

[tex]P(X=6)+P(X=7)\\=0.251[/tex]

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