The marginal cost of manufacturing an item when x thousand items are produced is dC/dx= 4x^3 - 6x + 5 dolars/item.Find the cost function C(x) if C(0)=550

Answers

Answer 1

Answer:

The cost function is [tex]C(x)=x^4-3x^2+5x+550[/tex].

Step-by-step explanation:

It is given that the marginal cost of manufacturing an item when x thousand items are produced is

[tex]\frac{dC}{dx}=4x^3-6x+5[/tex]

We need to find the cost function.

Multiply both sides by dx.

[tex]dC=(4x^3-6x+5)dx[/tex]

Integrate both sides to find the cost function.

[tex]\int dC=\int (4x^3-6x+5)dx[/tex]

[tex]\int dC=4\int x^3dx-6\int xdx+5\int 1 dx[/tex]

[tex]C(x)=4(\frac{x^4}{4})-6(\frac{x^2}{2})+5x+C[/tex]

where, C(x) is const function and C is a constant.

[tex]C(x)=x^4-3x^2+5x+C[/tex]

It is given that C(0)=550. Substitute x=0 in the above function.

[tex]C(0)=(0)^4-3(0)^2+5(0)+C[/tex]

[tex]C(0)=C[/tex]

[tex]550=C[/tex]

Therefore, the cost function is [tex]C(x)=x^4-3x^2+5x+550[/tex].


Related Questions

Using the logistic model f(x)=1501+9e−2x, evaluate the function at f(4). Round your answer to the nearest tenth.

Answers

Answer:

[tex]f(4) = 1501 + 9e^{-2*4} = 1501.00[/tex]

Step-by-step explanation:

f(4) is the value of f when x = 4.

We have that

[tex]f(x) = 1501 + 9e^{-2x}[/tex]

So

[tex]f(4) = 1501 + 9e^{-2*4} = 1501.00[/tex]

Answer:1501

Step-by-step explanation:

The given logistic model is expressed as

f(x)=1501+9e−2x

To evaluate the function at f(4), we would substitute x = 4 into the given logistic model. It becomes

f(4)=1501+9e−2 × 4

f(4)=1501+9e−8

Input 1501 plus 9 plus shift Ln plus -8 in a calculator. It becomes

1501 + 0.00302 = 1501.00302

Approximating to the nearest tenth, it becomes I501

A scientist was interested in studying if students political beliefs change as they go through college. Two hundred randomly selected students were asked before they entered college if they would consider themselves liberal or conservative. Four years later, the same two hundred students were asked if they would consider themselves, liberal or conservative. The scientist decided to perform McNemar's test. The data is below. What is the null hypothesis? After College Before College Liberal Conservative Liberal 80 15 Conservative 20 85

A. -0.85 or 0.85

B. -0.39 or 0.39

C. -9.75 or -9.75

D. 1.96 or -1.96

Answers

The null hypothesis in this considered experiment is: There is no change in their political beliefs as they go through college.

How to form the hypotheses?

There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.

Null hypothesis is the one which researchers try to disprove.

Here, it is specified that the scientist wants to study if students political beliefs change as they go through college. He wants to test if there are changes in the proportions of people who are liberal( or conservative).

Given that:

100 students before and after were asked their policial belief, as shown in table:

                               Liberal                Conservative

After college              80                       20

Before college           85                       15

Proportion of liberal = 1 - proportion of conservatives

So we will symbolize the hypotheses in terms of one of them, let it be proportions of liberals.

Sample size = 100

Favorable cases = Number of people from sample who consider themselves liberal. = X (say)

After college:

Sample proportion is: [tex]\hat{p}_1 = \dfrac{X}{N} = \dfrac{80}{100} = 0.8[/tex]

Before college:

Sample proportion is: [tex]\hat{p}_2 = \dfrac{X}{N} = \dfrac{85}{100} = 0.85[/tex]

Let p1 and p2 be the population proportion of people believing themself liberal after and before college respectively.

Then, the null hypothesis will assume that the claim of difference in belief the scientist wants to test is false, and therfore,

[tex]H_0: p_1 = p_2[/tex] or [tex]H_0: p_1 - p_2 = 0[/tex] (no difference, proportions are same, indicating that belief of students doesn't change much, as d)

Thus, the null hypothesis in this considered experiment is: There is no change in their political beliefs as they go through college.

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What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−23x+5 ?

Answers

Answer:

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

m = slope

The equation of the given line is

y = - 23x+5

Comparing it with the slope intercept equation, slope, m = -23

If a line is perpendicular to another line, the slope of the line is the negative reciprocal of the given line. This means that the slope of the line passing through the point (8, 1) is 1/23

We would determine the intercept, c by substituting m = 1/23, x = 8 and y = 1 into y = mx + c. It becomes

1 = 1/23 × 8 + c

1 = 8/23 + c

c = 1 - 8/23 = 15/23

The equation becomes

y = x/23 + 15/23

On the average, 1.6 customers per minute arrive at any one of the checkout counters of Sunshine food market.

What type of probability distribution can be used to find out the probability that there will be no customers arriving at a checkout counter in 10 minutes?
-Poisson distribution
-Normal distribution
-Binomial distribution
-None of these choices.

Answers

the answer would easily be Poisson distribution

The effects of a weight loss drug are standard normally distributed where negative data values represent weight loss. What is the probability a person loses 1.5 pounds or more? (round your answer to the nearest thousandth)My math:Weight loss of 1.5lbs, Z-score = -1.5Probability a person loses 1.5+ lbs = P(x > -1.5)1 – P(X > -1.5)1 - 0.0668 = 0.9332, or 0.932 (this answer was labeled as WRONG)Comment from online quiz: What is the z-score? How can you find the probability from the z-table?Please help clarify what I did wrong. Thanks! -Michelle

Answers

Answer:

[tex]P(X\leq -1.5) = P(X < -1.5)=P(Z<-1.5)=0.067[/tex]

Step-by-step explanation:

For this case we know that our random variable X="weight loss or gain" is distributed on this way:

[tex]X \sim N (\mu =0, \sigma=1)[/tex]

And we want the probability a person loses 1.5 pounds or more. If we interpret this an individual person losses 1.5 pounds or more if our random variable is equal or lower than 1.5. That means this:

[tex]P(X\leq -1.5) = P(X < -1.5)=P(Z<-1.5)[/tex]

And for this case we can use the normal standard distribution or excel with the following code:

"=NORM.DIST(-1.5,0,1,TRUE)"

And we got:

[tex]P(X\leq -1.5) = P(X < -1.5)=P(Z<-1.5)=0.067[/tex]

We need to remember that if the negative number decrease on the weight loss we are increasing the loss. For this reason we just need to find P(X<-1.5).

Complete the two column proof Given: 22= 24,m_2 = 110°
Prove: m 23 = 70°
Statement
Proof
22 24, mZ2=1109
Given
m 2 2 = m 24
Definition of congruent angles
m 24 = 1100
m 23 and m 24 are a linear pair
Definition of a linear pair (shown in
diagram)​

Answers

Step-by-step explanation:

I took 3 indicated as 5 and its adjacent angle to be 3

<2 = <4

As <2 =<3( corresponding angle)

And <3 = <4 ( Vertically opp.angle)

hence <2 = <4

<2 =>110

so , <4 = 110

So, <4 =>110

As <4 and <5 form linear pair

So <4 + <5 =>180

<5 = 180 -110 =>70

As i took <5 as replacing angle to <3

So According to Question fig

<3 =>70

Hence proved

A leprechaun places a magic penny under a girl's pillow. The next night there are 2 magic pennies under her pillow. Each night the number of magic pennies doubles. How much money will the girl have after 25 nights? PLEASE HURRY

Answers

Answer:

The girl will have $335,544.32

Step-by-step explanation:

2^25 = 33,554,432

Divide by 100 to turn the amount of pennies into dollars:

33,554,432/100

$335,544.32

the standard deviation of a set of 5 different integers each of which is between 0 and 10

Answers

Answer:

Mean = 3.8, std dev = 3.06

Step-by-step explanation:

First let us select any 5 integers without repitition from 0 to 10

Let this be 0,2,3,5,9

To calculate mean and standard deviation

x (x-3.8)^2  

   

0 14.44  

2 3.24  

3 0.64  

5 1.44  

9 27.04  

   

total 19 46.8  

   

Mean 3.8 9.36 Variance

   

Variance 9.36  

Std dev 3.059411708  

Please note that this mean and std deviation would vary according to our selection of 5 integers.

The standard deviation of the set {2, 4, 6, 8, 10} is approximately 2.83.

To calculate the standard deviation of a set of 5 different integers, each of which is between 0 and 10, you'll first need to find the mean (average) of the set and then calculate the squared differences from the mean. Here are the steps:

Find the mean (average) of the set:

Add up all the integers and divide by the number of integers (in this case, 5).

Calculate the squared differences from the mean:

For each integer in the set, subtract the mean and then square the result. Do this for all 5 integers.

Find the variance:

Add up all the squared differences from step 2 and divide by the number of integers (5 in this case).

Calculate the standard deviation:

Take the square root of the variance to find the standard deviation.

Let's go through an example:

Suppose you have the following set of 5 different integers between 0 and 10: {2, 4, 6, 8, 10}

Find the mean:

(2 + 4 + 6 + 8 + 10) / 5 = 30 / 5 = 6

Calculate the squared differences from the mean:

(2 - 6)^2 = 16

(4 - 6)^2 = 4

(6 - 6)^2 = 0

(8 - 6)^2 = 4

(10 - 6)^2 = 16

Find the variance:

(16 + 4 + 0 + 4 + 16) / 5 = 40 / 5 = 8

Calculate the standard deviation:

√8 ≈ 2.83 (rounded to two decimal places)

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What percentage of youth sport participants has experienced a sport related injury?

A.50%100%
B. 75%
C. 25%
D. 95%

Answers

Answer:

B.75%

Step-by-step explanation:

Answer:

A. 50%

Step-by-step explanation:

about 50% b/c over half the people who play sports all get injuries

It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the same among the 25 to 30 age group. Question 1. How many of this younger age group must we survey in order to estimate the proportion of non-grads to within .10 with 90% confidence? Use the value of p from the over-50 age group. (Round up to the nearest integer.) n = Question 2. Suppose we still want 90% confidence but we want to cut the margin of error to .04. What is the necessary sample size? (Round up to the nearest integer.) n = Question 3. What sample size is needed to estimate the proportion of non-grads to within .04 with 95% confidence? (Round up to the nearest integer.) n =

Answers

Answer:

1)  n=48  

2) n=298

3) n=426

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[/tex] represent the real population proportion of interest

[tex]\hat p[/tex] represent the estimated proportion for the sample

n is the sample size required (variable of interest)

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

The population proportion have the following distribution  

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

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by [tex]\alpha=1-0.90=0.10[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:  

[tex]z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64[/tex]  

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)  

And on this case we have that [tex]ME =\pm 0.1[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

[tex]n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63[/tex]  

And rounded up we have that n=48  

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

[tex]n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7[/tex]  

And rounded up we have that n=298  

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:  

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/tex]  

The margin of error for the proportion interval is given by this formula:  

[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)  

And on this case we have that [tex]ME =\pm 0.04[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:  

[tex]n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.96})^2}=425.22[/tex]  

And rounded up we have that n=426  

The probability shows that the number of younger age group that must will be surveyed will be 48.

How to compute the probability?

From the information given, the confidence level is 90%, margin or error is 0.1. The critical value from the z table is given as 1.645.

The number of samples will be:

= 0.23 × (1 - 0.23) × (1.645/0.1)²

= 0.23 × 0.77 × (1.645/0.1)²

= 48

The sample size when we are 90% confidence but we want to cut the margin of error to .04 will be computed thus:

= (1.645/.004)² × 0.23 × (1 - 0.23)

= (41.125)² × 0.1771

= 300

The sample size that is needed to estimate the proportion of non-grads to within .04 with 95% confidence will be:

= (1.96/0.04)² × 0.23 × (1 - 0.23)

= 48.99 × 0.1771

= 426

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Verify that P = Ce^t /1 + Ce^t is a one-parameter family of solutions to the differential equation dP dt = P(1 − P).

Answers

Answer:

See verification below

Step-by-step explanation:

We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that

[tex]1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}[/tex]

Now, differentiate to obtain

[tex]\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}[/tex]

[tex]=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}[/tex]

To obtain the required form, extract a factor in both the numerator and denominator:

[tex]\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)[/tex]

Assume that the weights of Chinook Salmon in the Columbia River are normally distributed. You randomly catch and weigh 40 such salmon. The mean weight from your sample is 23.6 pounds with a standard deviation of 3.5 pounds. Test the claim that the mean weight of Columbia River salmon is greater than 23 pounds. Test this claim at the 0.10 significance level.
(a) What type of test is this?i) This is a right-tailed test.ii) This is a two-tailed test. iii) This is a left-tailed test.(b) What is the test statistic? Round your answer to 2 decimal places.

Answers

Answer:

a) i) This is a right-tailed test.

b)

[tex]\text{Test statistic} = 1.08[/tex]

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 23 pounds

Sample mean, [tex]\bar{x}[/tex] = 23.6 pounds

Sample size, n = 40

Alpha, α = 0.10

Sample standard deviation, s = 3.5 pounds

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 23\text{ pounds}\\H_A: \mu > 23\text{ pounds}[/tex]

This is a one tailed(right).

Formula:

[tex]\text{Test statistic} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]\text{Test statistic} = \displaystyle\frac{23.6 - 23}{\frac{3.5}{\sqrt{40}} } = 1.08[/tex]

A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 56 men and 80 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained.Two sample T for Men vs WomenN Mean StDev SE MeanMen 56 112.5 11.1 1.5Women 80 118.7 14.2 1.695% CI for men – mu WomenT-Test mu Men = wu Women (vs<)T = 2.85 P = 0.0025 DF = 132State the null and alternative hypotheses. Which of the following is correct?a)A. H0: u1 = u2; Ha :u1>u2B. H0:u1 = u2;Ha::u1 not equal u2C. H0:u1 = u2;Ha:u1

Answers

Answer:

Null hypothesis:[tex]\mu_{m} \geq \mu_{w}[/tex]

Alternative hypothesis:[tex]\mu_{m} < \mu_{w}[/tex]

[tex]p_v =P(t_{134}<-2.85)=0.0025[/tex]

D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.

So on this case the 95% confidence interval would be given by [tex]-10.502 \leq \mu_{m} -\mu_w \leq -1.898[/tex]  

We are 95% confident that the mean difference is in the confidence interval.

Step-by-step explanation:

1) Data given and notation

[tex]\bar X_{m}=112.5[/tex] represent the mean for the sample of men

[tex]\bar X_{w}=118.7[/tex] represent the mean for the sample women

[tex]s_{m}=11.1[/tex] represent the sample standard deviation for the sample men

[tex]s_{w}=14.2[/tex] represent the sample standard deviation for the sample eomen

[tex]n_{m}=56[/tex] sample size for the group men

[tex]n_{w}=80[/tex] sample size for the group women

z would represent the statistic (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the means for the two groups are the same, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{m} \geq \mu_{w}[/tex]

Alternative hypothesis:[tex]\mu_{m} < \mu_{w}[/tex]

We don't have the population standard deviation's, so for this case is better apply a t test to compare means, and the statistic is given by:

[tex]t=\frac{\bar X_{m}-\bar X_{w}}{\sqrt{\frac{s^2_{m}}{n_{m}}+\frac{s^2_{w}}{n_{w}}}}[/tex] (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

With the info given we can replace in formula (1) like this:

[tex]t=\frac{112.5-118.7}{\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}}=-2.85[/tex]  

4) Statistical decision

The degrees of freedom are given by:

[tex]df=n_m +n_w -2= 56+80-2=134[/tex]

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

[tex]p_v =P(t_{134}<-2.85)=0.0025[/tex]

D. Reject H0, there is sufficient evidence to conclude that the mean step pulse of men was less than the mean step pulse of women.

5) Confidence interval

The confidence interval for the difference of means is given by the following formula:  

[tex](\bar X_m -\bar X_w) \pm t_{\alpha/2}\sqrt{(\frac{s^2_m}{n_m}+\frac{s^2_w}{n_w})}[/tex] (1)  

The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:  

[tex]\bar X_m -\bar X_w =112.5-118.7=-6.2[/tex]  

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,134)".And we see that [tex]z_{\alpha/2}=1.98[/tex]  

Now we have everything in order to replace into formula (1):  

[tex]-6.2-1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-10.502[/tex]  

[tex]-6.2+1.98\sqrt{\frac{11.1^2}{56}+\frac{14.2^2}{80}}}=-1.898[/tex]  

So on this case the 95% confidence interval would be given by [tex]-10.502 \leq \mu_{m} -\mu_w \leq -1.898[/tex]  

We are 95% confident that the mean difference is in the confidence interval.

Final answer:

In the context of the study conducted by the physical therapist, the null hypothesis states that the mean step pulse of men equals the mean step pulse of women, and the alternative hypothesis states that the mean step pulse of men is less than that of women. With a t-test result that has a P-value less than 0.05, we reject the null hypothesis and accept the alternative hypothesis, indicating a significant difference between the mean step pulses of men and women.

Explanation:

The physical therapist's study is essentially a hypothesis testing problem. The null hypothesis (σ_0) is that the mean step pulse of men is equal to the mean step pulse of women, and the alternative hypothesis (σ_a) is that the mean step pulse of men is less than the mean step pulse of women.

For this problem, we can write the hypotheses as follows:
σ_0: μ1 = μ2
σ_a: μ1 < μ2

The student's available choices seem to reflect this. Choosing between 'H0: μ1 = μ2; Ha :μ1>μ2' and 'H0:μ1 = μ2;Ha::μ1 not equal μ2'. The correct choice is not listed because the alternative hypothesis is incorrectly defined in both cases. The correct alternative hypothesis would be 'μ1 < μ2', pointing that the mean step pulse of men is less than that of women.

The t-test result given in the question 'T = 2.85 P = 0.0025 DF = 132' indicates that the mean step pulse of men is not equal to that of women, with the P-value being less than 0.05. Thus, with this information, you would typically reject the null hypothesis and accept the alternative hypothesis, which suggests a significant difference between the mean step pulses of men and women.

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Audrey, an astronomer is searching for extra-solar planets using the technique of relativistic lensing. Though there are believed to be a very large number of planets that can be found this way, actually finding one takes time and luck; and finding one planet does not help at all with finding planets of other stars in the same part of the sky. Audrey is good at it, and finds one planet at a time, on average once every three months.

a.) Find the expected value and standard deviation of the number of planets she will find in the next two years.
b.) When she finds her sixth new planet, she will be eligible for a prize. Find the expected value and standard deviation of the amount of time until she is eligible for that prize.
c.) Find the probability that she will become eligible for that prize within one year.

Answers

Answer:

Step-by-step explanation:

The model [tex]N (t)[/tex], the number of planets found up to time [tex]t[/tex], as a Poisson process. So, the [tex]N (t)[/tex] has distribution of Poison distribution with parameter [tex](\lambda t)[/tex]

a)

The mean for a month is, [tex]\lambda = \frac{1}{3}[/tex] per month

[tex]E[N(t)]= \lambda t\\\\=\frac{1}{3}

(24)\\\\=8[/tex]

(Here. t = 24)

For Poisson process mean and variance are same,

[tex]Var[N (t)]= Var[N(24)]\\= E [N (24)]\\=8[/tex]

 

(Poisson distribution mean and variance equal)

 

The standard deviation of the number of planets is,

[tex]\sigma( 24 )] =\sqrt{Var[ N(24)]}=\sqrt{8}= 2.828

[/tex]

b)

For the Poisson process the intervals between events(finding a new planet) have  independent  exponential  distribution with parameter [tex]\lambda[/tex]. The  sum  of [tex]K[/tex] of these  independent exponential has distribution Gamma [tex](K, \lambda)[/tex].

From the given information, [tex]k = 6[/tex] and [tex]\lambda =\frac{1}{3}

[/tex]

Calculate the expected value.

[tex]E(x)=\frac{\alpha}{\beta}\\\\=\frac{K}{\lambda}

\\\\=\frac{6}{\frac{1}{3}}\\\\=18[/tex]

(Here, [tex]\alpha =k[/tex] and [tex]\beta=\lambda[/tex])                                                                      

C)

Calculate the probability that she will become eligible for the prize within one year.

Here, 1 year is equal to 12 months.

P(X ≤ 12) = (1/Г  (k)λ^k)(x)^(k-1).(e)^(-x/λ)

[tex]=\frac{1}{Г  (6)(\frac{1}{3})^6}(12)^{6-1}e^{-36}\\\\=0.2148696\\=0.2419\\=21.49%[/tex]

Hence, the required probability is 0.2149 or 21.49%

Final answer:

We can calculate the expected value and standard deviation of the number of planets Audrey will find in the next two years using probability and statistics. Using geometric distribution, we can also calculate the expected value and standard deviation of the amount of time until Audrey is eligible for a prize after finding her sixth new planet. Finally, we can calculate the probability of Audrey becoming eligible for the prize within one year using the binomial distribution formula.

Explanation:

To find the expected value and standard deviation of the number of planets Audrey will find in the next two years, we can use the concepts of probability and statistics. Since she finds one planet every three months on average, the expected value can be calculated by multiplying the average number of planets found per month (1/3) by the number of months in two years (24). The standard deviation can be calculated using the formula sqrt(n * p * q), where n is the number of trials, p is the probability of success, and q is the probability of failure.

To find the expected value and standard deviation of the amount of time until Audrey is eligible for the prize after finding her sixth new planet, we can use the concept of geometric distribution. The expected value can be calculated by taking the reciprocal of the probability of success (1/6) and the standard deviation can be calculated using the formula sqrt((1-p) / (p^2)).

To find the probability that Audrey will become eligible for the prize within one year, we can calculate the cumulative probability of finding six or more new planets in one year using the binomial distribution formula.

Learn more about Expected value and standard deviation here:

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One study reports that 34​% of newly hired MBAs are confronted with unethical business practices during their first year of employment. One business school dean wondered if her MBA graduates had similar experiences. She surveyed recent graduates from her​ school's MBA program to find that 28​% of the 116 graduates from the previous year claim to have encountered unethical business practices in the workplace. Can she conclude that her​ graduates' experiences are​ different?

Answers

Answer:

[tex]z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364[/tex]  

[tex]p_v =2*P(Z<-1.364)=0.173[/tex]  

The p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.  

Step-by-step explanation:

1) Data given and notation

n=116 represent the random sample taken

X represent the number graduates from the previous year claim to have encountered unethical business practices in the workplace

[tex]\hat p=0.28[/tex] estimated proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace

[tex]p_o=0.34[/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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is 0.34 or no.:  

Null hypothesis:[tex]p=0.34[/tex]  

Alternative hypothesis:[tex]p \neq 0.34[/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].

3) Calculate the statistic  

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

[tex]z=\frac{0.28 -0.34}{\sqrt{\frac{0.34(1-0.34)}{116}}}=-1.364[/tex]  

4) 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 bilateral test the p value would be:  

[tex]p_v =2*P(Z<-1.364)=0.173[/tex]  

The p value obtained was a very high value and using the significance level assumed [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIl to reject the null hypothesis, and we can said that at 5% of significance the proportion of graduates from the previous year claim to have encountered unethical business practices in the workplace is not significant different from 0.34.  

Management at a seaside resort is publishing a brochure and wants to include a statement about the proportion of clear days during their peak season. Out of a random sample of 150 days from over the last two peak seasons, 117 days were recorded as clear. They want to estimate the proportion of clear days to within a 5% margin of error with a 95% confidence interval. What's the sample size necessary to construct this interval?A. 384B. 264C. 383D. 385E. 263

Answers

Answer: B. 264

Step-by-step explanation:

Formula to calculate the sample size 'n' , if the prior estimate of the population proportion (p) is available:

[tex]n= p(1-p)(\dfrac{z}{E})^2[/tex]

, where z = Critical z-value corresponds to the given confidence interval

E=  margin of error

Let p be the population proportion of clear days.

As per given , we have

Prior sample size : n= 150

Number of clear days in that sample = 117

Prior estimate of the population proportion of clear days = [tex]p=\dfrac{117}{150}[/tex]

E= 0.05

The critical z-value corresponding to 95% confidence interval = z*= 1.95 (By z-table)

Then, the required sample size will be :

[tex]n= \dfrac{117}{150}(1-\dfrac{117}{150})(\dfrac{1.96}{0.05})^2[/tex]

Simplify ,

[tex]n= (0.1716)(39.2)^2[/tex]

[tex]n= 263.687424\approx264[/tex]

Hence, the sample size necessary to construct this interval =264

Thus the correct option is B. 264

Robin can clean 72 rooms in 666 days.
How many rooms can Robin clean in 999 days?
Will give brainiest to correct answer!!!!!!!!!!!!!!

Answers

Answer:

108 rooms

Step-by-step explanation:

Robin can clean 72 rooms in 666 days

means in 666 days rooms that can be clean  is 72 rooms

in 1 days rooms that can be clean  is 72/666 rooms

in 999 days rooms can be clean is  999\times 72\div 666

after solving we get 108 answer

A random sample of the amounts for 22 purchases was taken. The mean was ​$42.97​, the standard deviation was ​$22.82​, and the margin of error for a 95​% confidence interval was ​$10.12. Assume that t Subscript n minus 1 Superscript starequals2.0 for the​ 95% confidence intervals. ​a) To reduce the margin of error to about ​$5​, how large would the sample size have to​ be? ​b) How large would the sample size have to be to reduce the margin of error to ​$1.0​?

Answers

Answer: a) 84 and b ) 2084

Step-by-step explanation:

Given : Sample standard deviation : s= $22.82

(Population standard deviation is unknown ) , so we use t-test.

Critical value or the​ 95% confidence intervals :[tex]t_{n-1}*=2.0[/tex]

Formula to find the sample size :

[tex]n=(\dfrac{t^*\cdot s}{E})^2[/tex]

a) E = 5

[tex]n=(\dfrac{(2)\cdot 22.82}{5})^2[/tex]

[tex]n=(9.128)^2=83.320384\approx84[/tex]

i.e. Required sample size : n= 84

b)  E = 1

[tex]n=(\dfrac{(2)\cdot 22.82}{1})^2[/tex]

[tex]n=(45.64)^2=2083.0096\approx2084[/tex]

i.e. Required sample size : n= 2084

The side of the base of a square prism is increasing at a rate of 5 meters per second and the height of the prism as decreasing at a rate of 2 meters per second. At a certain instant, the base's side is 6 meters and the height is 7 meters. What is the rate of change of the volume of the prism at that instant fin cubic meters per second?

A. -348
B. 492
C. -492
D. 348

Answers

Answer:

D. 348

Step-by-step explanation:

The volume of the square prisma is given by the following formula:

[tex]V = s^{2}h[/tex]

In which h is the height, and s is the side of the base.

Let's use implicit derivatives to solve this problem:

[tex]\frac{dV}{dt} = 2sh\frac{ds}{dt} + s^{2}\frac{dh}{dt}[/tex]

In this problem, we have that:

[tex]\frac{ds}{dt} = 5, \frac{dh}{dt} = -2, h = 7, s = 6[/tex]

So

[tex]\frac{dV}{dt} = 2sh\frac{ds}{dt} + s^{2}\frac{dh}{dt}[/tex]

[tex]\frac{dV}{dt} = 2*6*7*5 + (6)^{2}*(-2) = 348[/tex]

So the correct answer is:

D. 348

Use this information when answering this question:

n = 14

s = 20

H0: σ2 ≤ 500 and Ha: σ2 > 500

The null hypothesis:
a. should be revised
b. should not be rejected
c. should be rejected
d. None of these alternatives is correct.

Answers

Answer:   c. should be rejected  

Step-by-step explanation:

n = 14

s = 20

H0: σ2 ≤ 500 and Ha: σ2 > 500

Use technology to find the​ P-value for the hypothesis test described below. The claim is that for 12 AM body​ temperatures, the mean is mu μ greater than > 98.6 98.6 degrees °F. The sample size is n equals = 9 9 and the test statistic is t equals = 2.132 2.132.

Answers

Answer:

P-value = 0.032794

Step-by-step explanation:

We are given the following information in the question:

Population mean, μ = 98.6 degrees

Sample size, n = 9

Alpha, α = 0.05

Test t-statistic = 2.132

The null and the alternate hypothesis :

[tex]H_{0}: \mu = 98.6\text{ degrees}\\H_A: \mu > 98.6\text{ degrees}[/tex]

We have to find the p-value for degree of freedom 8 and significance level 0.05

The calculated p-value is 0.032794

Jhons garden has a big planter that mensures 18 2/3 in by 8 5/6 in. What Is the area of Jhons planter

Answers

Area of Jhons planter is 164.89 square inches

Solution:

Given that,

Jhons garden has a big planter that measures [tex]18\frac{2}{3}[/tex] in by [tex]8 \frac{5}{6}[/tex] inches

To find: area of Jhons planter

From given information,

[tex]\text{ length } = 18\frac{2}{3} = \frac{3 \times 18 + 2}{3} = \frac{56}{3} \text{ inches }[/tex]

[tex]\text{ width } = 8\frac{5}{6} = \frac{6 \times 8 + 5}{6} = \frac{53}{6} \text{ inches }[/tex]

The area of planter is given as:

[tex]area = length \times width[/tex]

[tex]area = \frac{56}{3} \times \frac{53}{6} = \frac{2968}{18} = 164.89[/tex]

Thus area of Jhons planter is 164.89 square inches

The radioactive isotope carbon-14 is present in small quantities in all life forms, and it is constantly replenished until the organism dies, after which it decays to stable carbon-12 at a rate proportional to the amount of carbon-14 present, with a half-life of 5595 years. Suppose C(t) is the amount of carbon-14 present at time t.(a) Find the value of the constant k in the differential equation.(b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained about 91 percent of the amount of carbon-14 contained in freshly made cloth of the same material. How old is the Shroud of Turin, according to these data?

Answers

Answer:

a) [tex]k = 0.000124[/tex]

b) According to these data, the Shroud of Turin has around 760 years.

Step-by-step explanation:

The amount of carbon-14 is modeled by the following equation:

[tex]C(t) = C_{0}e^{-kt}[/tex]

In which [tex]C_{0}[/tex] is the initial amount and k is the rate of decrease.

(a) Find the value of the constant k in the differential equation.

Half-life of 5595 years.

So [tex]C(5595) = 0.5C_{0}[/tex]

[tex]C(t) = C_{0}e^{-kt}[/tex]

[tex]0.5C_{0} = C_{0}e^{-5595k}[/tex]

[tex]e^{-5595k} = 0.5[/tex]

Applying ln to both sides

[tex]-5595k = -0.69[/tex]

[tex]k = 0.000124[/tex]

b) In 1988 three teams of scientists found that the Shroud of Turin, which was reputed to be the burial cloth of Jesus, contained about 91 percent of the amount of carbon-14 contained in freshly made cloth of the same material. How old is the Shroud of Turin, according to these data?

This is t when [tex]C(t) = 0.91C_{0}[/tex]

[tex]C(t) = C_{0}e^{-kt}[/tex]

[tex]0.91C_{0} = C_{0}e^{-0.000124t}[/tex]

[tex]e^{-0.000124t} = 0.91[/tex]

Applying ln to both sides

[tex]-0.000124t = -0.094[/tex]

[tex]t = 760.57[/tex]

According to these data, the Shroud of Turin has around 760 years.

From the top of a vertical tower, 331 feet above the surface of the earth, the angle of depression to a doghouse is 28 degrees 8'. How far is it from the doghouse to the foot of the tower?

Answers

Answer:

619.13 feet

Step-by-step explanation:

Please find the attachment.

Let x represent the distance between doghouse to the foot of the tower.

We have been given that from the top of a vertical tower, 331 feet above the surface of the earth, the angle of depression to a doghouse is 28 degrees 8'. We are asked to find the distance between doghouse to the foot of the tower.

First of all, we will convert our given angle into degrees as it is given in degrees and minutes.

We will divide 8 by 60 to convert 8 minutes into degrees as:

[tex]\frac{8}{60}=0.13333[/tex]

The doghouse, tower and angle of depression forms a right triangle with respect to ground, where, 331 feet is opposite side and x is adjacent side to angle 28.13 degrees.

[tex]\text{tan}=\frac{\text{Opposite}}{\text{Adjacent}}[/tex]

[tex]\text{tan}(28.13^{\circ})=\frac{331}{x}[/tex]

[tex]x=\frac{331}{\text{tan}(28.13^{\circ})}[/tex]

[tex]x=\frac{331}{0.534623339864}[/tex]

[tex]x=619.12747783[/tex]

Upon rounding to nearest hundredth, we will get:

[tex]x\approx 619.13[/tex]

Therefore, the doghouse is 619.13 feet far from the foot of the tower.

Help please!!! 20 points will mark brainliest!! :)

Answers

Answer:

The equation has zeroes at -4, 0 and 4 and is a minimum cubic degree or a 3 degree equation.

Step-by-step explanation:

The polynomial has roots at places where f(x) cuts the x axis.

The function cuts the x-axis at 3 points: 4, -4, and 0.

It cuts the axis at the point, x=0; x=4; x=-4

Therefore, the points equation has to be of the form, k*x*(x-4)*(x+4)*p(x)

where k is any arbitrary constant and p(x) is a polynomial of any degree depending on what the equation does in the region and (5,∞) and (-∞,-5).

Therefore, the equation has zeroes at -4, 0 and 4 and is a minimum cubic degree or a 3 degree equation.

Two hundred randomly selected people were asked to state their primary source for news about current events: (1) Television, (2) Radio, (3) Internet, or (4) Other. We wish to determine whether the percent distribution of responses is 10%, 30%, 50%, and 10% for news sources 1 through 4, respectively. What is the null hypothesis? a. H0: p1= .10, p2 = .30, p3 = .50, p4 = .10 b. H0: p1 = .25, p2 = .25, p3 = .25, p4 = .25 c. H0: p = .50 d. H0: the proportions are not all equal.

Answers

Answer:

H0 as the proportions are not as per the given estimation

Step-by-step explanation:

Given that two hundred randomly selected people were asked to state their primary source for news about current events: (1) Television, (2) Radio, (3) Internet, or (4) Other.

The hypothesised distribution is with 10%, 30%, 50%, and 10% for news sources 1 through 4, respectively

To check whether this is correct, we need not do test for each proportion

Instead we can do hypothesis testing for chi square goodness of fit with

H0 as the proportions are not as per the given estimation

You test calories for a food item. The brand name has a mean of 158.706 and a sample standard deviation = 25.236, when seventeen are tested. The generic item has a mean of 122.471 and a sample standard deviation = 25.183, when seventeen are tested. Which is a confdence interval of 95%?

Multiple Choice: (Show work)

A) 17.21 to 55.26
B) 18.12 to 54.35
C) 17.79 to 54.67
D) 18.622 to 53.848

Answers

Answer:

option C

Step-by-step explanation:

given,

[tex]\bar{x_1} = 158.706, \sigma_1 = 25.36 , n_1= 17[/tex]

[tex]\bar{x_1} = 122.471, \sigma_1 = 25.183 , n_2= 17[/tex]

α = 1 - 0.95 = 0.05

degree of freedom (df) = 17 -1 = 16

critical value[tex]= t_{\alpha/2},df = t_{0.025},16 =2.120[/tex] (from t-table)

margin of error = [tex]t_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}[/tex]

=2.120\times \sqrt{\dfrac{25.36^2}{17}+\dfrac{25.183^2}{17}}[/tex]

= 2.120 x 8.6467

= 18.33

Margin of error = 18.33

Point estimation of difference = [tex]\bar{x_1} - \bar{x_2}[/tex]

                                                  = 36.235

lower limit = 36.235 - 18.33 = 17.91

upper limit = 36.235 + 18.33 = 54.57

hence, the nearest option near to answer is option C

Considered safe for agricultural use. A well in Texas is used to water crops. This well is tested on a regular basis for arsenic. A random sample of 36 tests gave a sample mean of = 7.3 ppb arsenic, with s = 1.9 ppb. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use a 0.01 level of signifcance..

Answers

Answer:

This information indicates that the mean level of arsenic in this well is less than 8 ppbat  0.01 level of signifcance..

Step-by-step explanation:

Given that a well in Texas is used to water crops.

This well is tested on a regular basis for arsenic.

A random sample of 36 tests gave a sample mean of = 7.3 ppb arsenic, with s = 1.9 ppb

H0: Sample mean = 8

Ha: Sample mean <8

(left tailed test at 1% level)

Mean difference =-0.70

Std error of mean = [tex]\frac{1.9}{\sqrt{36} } \\=0.3167[/tex]

Test statistic t = -2.204

df = 35

p value = 0.0171

Since p >0.01 we accept H0

This information indicates that the mean level of arsenic in this well is less than 8 ppbat  0.01 level of signifcance..

What is 6 to the power of 8 in exponential form?

Answers

Answer: 1.7 x 10^6

Step-by-step explanation:

6^8 = 1,679,616

1,679,616 = 1.7 x 10^6

Need Help ASAP!! 100 points andbranliest plz pleople plz help.

Answers

Answer:

what grade are you in so then I could get the answer

Answer:

[tex] |78| + |12| = |12| + |78| [/tex]

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