......Help Please......

Answer:

The standard error increases and the width of the interval increases

Step-by-step explanation:

The variance is equal to the square of the standard deviation. An increase in variance will increase the standard deviation.

The width of the confidence interval can be defined as:

[tex]W=UL-LL=2z\sigma/\sqrt{n}[/tex]

The width is proportional to the standard deviation, so when the variance increases, the standard deviation and also the width of the interval increases.

A financial analyst wanted to estimate the mean annual return on mutual funds. A random sample of 60 funds' returns shows an average rate of 12%. If the population standard deviation is assumed to be 4%, the 95% confidence interval estimate for the annual return on all mutual funds is

A. 0.037773 to 0.202227

B. 3.7773% to 20.2227%

C. 59.98786% to 61.01214%

D. 51.7773% to 68.2227%

E. 10.988% to 13.012%

Answer: E. 10.988% to 13.012%

Step-by-step explanation:

Given;

Mean x= 12%

Standard deviation r = 4%

Number of samples tested n = 60

Confidence interval is 95%

Z' = t(0.025)= 1.96

Confidence interval = x +/- Z'(r/√n)

= 12% +/- 1.96(4%/√60)

= 12% +/- 0.01214%

Confidence interval= (10.988% to 13.012%)

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:

We accept  H₀   with the information we have, we can say level of ozone is under the major limit

Step-by-step explanation:

Normal Distribution

population mean  =   μ₀   =  7.5   ppm

Sample size     n   =  16      df  =  n  -  1  df  =  15

Sample mean      =    μ   =  7.8  ppm

Sample standard deviation   =  s  =  0.8

We want to find out if ozono level, is above normal level that is bigger than 7.5

1.- Hypothesis Test

null hypothesis                          H₀         μ₀   =  7.5

alternative hypothesis             Hₐ          μ₀  >  7.5  

2.-Significance level    α  =  0.01   we will develop one tail-test (right)

then   for   df  =  15    and    α =  0,01   from t -student table we get

t(c)  = 2.624

3.-Compute  t(s)

t(s)  =  (  μ -  μ₀ ) / s /√n            ⇒   t(s)  = ( 7.8  -  7.5 )*4/0.8

t(s)  = 0.3*4/0.8

t(s)  = 1.5

4.-Compare   t(s)   and  t(c)

t(s)  <  t(c)       1.5  <  2.64

Then  t(s) is inside the acceptance region. We accept  H₀

Answer:

See the explanation

Step-by-step explanation:

(a)  

H0: Consultant with more experience has the higher population mean service rating.

H1: Consultant with more experience doesn't have the higher population mean service rating.

(b)  

t = 1.9923 (see the attached image)

(c)

The degrees of freedom for the test statistic,

df = 16

The P-value of the one tailed t- test with 16 degrees of freedom is,

P−value = tdist(X,df,tails)

P-value = tdist(1.9923,16,1)

P-value = 0.032

(d)​

Since, P-value 0.032 is less than the significance level 0.05, there is an enough evidence to reject the null hypothesis.

Hence, there is a sufficient evidence to conclude that Consultant with more experience doesn't have the higher population mean service rating.

Hope this helps!

Answer: D. 0.306

Step-by-step explanation:

Assuming a normal distribution for the annual salary for intermediate level executives, the formula for normal distribution is expressed as

z = (x - u)/s

Where

x = annual salary for intermediate level executives

u = mean annual salary

s = standard deviation

From the information given,

u = $74000

s = $2500

We want to find the probability that the mean annual salary of the sample is between $71000 and $73500. It is expressed as

P(71000 lesser than or equal to x lesser than or equal to 73500)

For x = 71000,

z = (71000 - 74000)/2500 = - 1.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.1151

For x = 73500,

z = (73500 - 74000)/2500 = - 0.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.4207

P(71000 lesser than or equal to x lesser than or equal to 73500) is

0.4207 - 0.1151 = 0.306

The 1.125 in the exponential function V(t) = 30,000(1.125)^t represents the growth factor of the investment, indicating an annual growth of 12.5%.

In the exponential function V(t) = 30,000(1.125)^t, the 1.125 represents the growth factor of the investment. This means each year, the value of the investment grows by 12.5%. It's the rate at which the initial value of the investment, 30,000 in this case, increases on an annual basis. Therefore, the exponential function shows us how the value of the investment changes over time with this constant growth factored in.

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Answer:

The equation of the line using function notation is

[tex]y=f(x)=-\frac{1}{4}x[/tex]

Step-by-step explanation:

Given points are (-20,5) and (-36, 9)

Now to find the equation of the line passes through these points

Let [tex](x_{1},y_{1})[/tex] and  [tex](x_{2},y_{2})[/tex] be the two given points  (-20,5) and (-36, 9) respectively.

To find slope

[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]

[tex]m=\frac{9-5}{-36-(-20)}[/tex]

[tex]m=\frac{4}{-36+20}[/tex]

[tex]m=\frac{4}{-16}[/tex]

[tex]m=-\frac{1}{4}[/tex]

Therefore  [tex]m=-\frac{1}{4}[/tex]

The equation of the line is of thr form y=mx+c

The point (-20,5) passes through the above line and  [tex]m=-\frac{1}{4}[/tex]

[tex]5=-\frac{1}{4}(-20)+c[/tex]

[tex]5=5+c[/tex]

[tex]c=0[/tex]

[tex]y=-\frac{1}{4}x+0[/tex]

Therefore  [tex]y=-\frac{1}{4}x[/tex]

Therefore the equation of the line using function notation is

[tex]y=f(x)=-\frac{1}{4}x[/tex]

Answer: 482

Step-by-step explanation:

Formula to find the sample size is given by :-

[tex]n= (\dfrac{z^*\times \sigma}{E})^2[/tex]                                    (1)

, where z* = critical z-value (two tailed).  

[tex]\sigma[/tex] = Population standard deviation and E = Margin of error.

As per given , we have

Margin of error : E= 3

[tex]\sigma=40[/tex]

Confidence level = 90%

Significance level =[tex]\alpha=1-0.90=0.10[/tex]

Using z-table , the critical value for 90% confidence=[tex]z^*=z_{\alpha/2}=z_{0.05}=1.645[/tex]

Required minimum sample size = [tex]n= (\dfrac{(1.645)\times (40)}{3})^2[/tex]   [Substitute the values in formula (1)]

[tex]n=(21.9333333333)^2[/tex]

[tex]n=481.07111111\approx482[/tex]  [ Round to the next integer]

Hence, the number of observations required is closest to 482.

Answer:

(C) A Type I error would be incorrectly failing to convict the defendant when he is guilty. A Type II error would be incorrectly convicting the defendant when he is innocent.

Step-by-step explanation:

Type I error is rejecting the true null hypothesis and type II error is not rejecting the false null hypothesis. Hence in this scenario, it will be:

A Type I error would be incorrectly convicting the defendant when he is innocent. A Type II error would be incorrectly failing to convict the defendant when he is guilty.

Option C is correct.

Answer:

110

Step-by-step explanation:

%20 increase = [tex]\frac{20}{100} 1300 = 260\\[/tex]

%50 decrease = [tex]\frac{50}{100}300 = 150[/tex]

%30 no change = [tex]\frac{30}{100}0 = 0[/tex]

Expected change = Increase - Decrease + No Change

Expected change = 260 - 150 + 0

Expected change = 110

Answer

The answer and procedures of the exercise are attached in the following archives.

The information complete of the exercise is attached in the sheet.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:

A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

Test statistic(Z) = 2.31

P-value = 0.0104

Step-by-step explanation:

Step 1

Null hypothesis: The average damaging temperature of nine iPods is 5°F

Alternate hypothesis: The average damaging temperature of nine iPods differs from 5°F

Step 2

Mean=5°F, Sd=3, df=n-1=9-1=8

The t-value corresponding to 8 degrees of freedom and 95% confidence level (5% significance level) is 2.306

Confidence Interval(CI) = (mean + or - t×sd/√n)

CI = (5 + 2.306×3/√8) = 5 + 6.918/2.828= 5+2.45=7.45°F

CI = (5 - 2.306×3/√8) = 5-2.45 = 2.55°F

Z = (sample mean - population mean)/(sd÷√n) = (5-2.55)/(3÷√8) = 2.45/1.061 = 2.31

Step 3

Using the standard distribution table, the cumulative area to the left of Z = 2.31 is 0.9896

P-value = 1 - 0.9896 = 0.0104

Step 4

Conclusion: A 95% confidence interval for the true mean minimum temperature that will damage an iPod is between 2.55°F and 7.45°F

Answer:

a) [0.278; 0.342]

b) No, the population the sample comes from may not be representative of college students, violating the Fundamental Rule for Using Data For Inference.

Step-by-step explanation:

Hello!

a)

The study variable is

X: Number of American adults that said that they do not get enough sleep each night.

The sample is n= 935 American Adults

Sample proportion 'p= 0.31

The margin of error of the 95% CI d= 0.032

The formula for a CI for the population proportion is:

['p ± [tex]Z_{1-\alpha /2}[/tex]*[tex]\sqrt{\frac{'p(1-'p)}{n} }[/tex]]

The basic structure of the CI formula is "Point estimator" ± "Margin of error"

So with the given data, you can calculate the CI.

[0.31 ± 0.032]

[0.278; 0.342]

So with a confidence level of 95%, you'd expect that the interval [0.278; 0.342] will contain the population proportion of the American adults that don't get enough sleep at night.

b)

The answer is no. The sample is of "American adults" there is no information on their education level, therefore it is possible that all of them have a college education or that none of the have. Since no further information about their education then you cannot be certain that this sample will be representative of the American College Students.

Then the correct option is B

I hope you have a SUPER day!

Answer:

a.  [tex]\displaystyle AL = \int\limits^5_2 {\sqrt{1+ \frac{9}{x^2}} \, dx[/tex]

b.  [tex]\displaystyle AL = 4.10322[/tex]

General Formulas and Concepts:

Calculus

Differentiation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Integration

Integration Property [Multiplied Constant]:                                                         [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:                                                       [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

U-Substitution

Arc Length Formula [Rectangular]:                                                                       [tex]\displaystyle AL = \int\limits^b_a {\sqrt{1+ [f'(x)]^2}} \, dx[/tex]

Step-by-step explanation:

Step 1: Define

Identify

y = 3ln(x)

Interval [2, 5]

Step 2: Find Arc Length

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Applications of Integration

Book: College Calculus 10e

The arc length of the curve [tex]\( y = 3 \ln x \)[/tex] over the interval [2, 5] is approximately 12.092 units, obtained through numerical methods such as Simpson's rule or a CAS like Wolfram Alpha.

a. Writing and Simplifying the Integral:

The arc length L of the curve y = f(x) over the interval [a, b] is given by the following definite integral:

L = [tex]\int_a^b \sqrt{1 + \left( \dfrac{dy}{dx} \right)^2} dx[/tex]

First, find the derivative dy/dx of the given function:

y' = [tex]\dfrac{d}{dx} (3 \ln x) = \dfrac{3}{x}[/tex]

Now, plug f(x) = 3 \ln x and [tex]f'(x) = \dfrac{3}{x}[/tex] into the arc length formula for the interval [2, 5]:

L = [tex]\int_2^5 \sqrt{1 + \left( \dfrac{3}{x} \right)^2} dx[/tex]

L = [tex]\int_2^5 \sqrt{1 + \dfrac{9}{x^2}} dx[/tex]

b. Evaluating or Approximating the Integral:

This integral cannot be solved analytically using standard techniques. Therefore, we can use numerical methods like Simpson's rule or a CAS (Computer Algebra System) to approximate the value.

Using Simpson's rule with 10 subintervals, we get an approximate arc length of:

L ≈ 12.092

Alternatively, using a CAS like Wolfram Alpha, we can obtain a more precise numerical value:

L ≈ 12.091971077

Therefore, the arc length of the curve is approximately 12.092 units.

Answer:

[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]

Step-by-step explanation:

probability distribution fucntion is given as

[tex] F_x = P(X \leq x)[/tex]

       [tex] = \frac{x -a}{b -a}      a<x< b[/tex]

where a indicate lower end estimate = $250,000

b indicate high end estimate = $350,000

probability greater than $337500

[tex] P(X \geq 337,500) = 1- P(X < 337500)[/tex]

                               [tex] = 1 - \frac{x -a}{b -a}[/tex]

                               [tex] = 1 - \frac{337500 - 250000}{350000 - 250000}[/tex]

[tex]P(X \geq 337,500) = 0.125 = 12.5%[/tex]

Calculate the probability of net income being greater than or equal to $337,500 for Excellence Corporation following a continuous uniform distribution. Probability will be equal to 12.5%.

The probability that net income will be greater than or equal to $337,500 is 0.25. To calculate this probability for a continuous uniform distribution, we need to find the proportion of the area under the distribution curve that corresponds to values greater than $337,500.

Step-by-step calculation:

Calculate the total range of net income: $350,000 - $250,000 = $100,000

Calculate the proportion of the area for values greater than $337,500: ($350,000 - $337,500) / $100,000 = 0.125

Since the distribution is continuous and uniform, the probability is equal to the proportion calculated: 0.125 = 12.5%

Answer:

Earth's average distance from the Sun is [tex]375[/tex] times more than is from the Moon.

Step-by-step explanation:

Given the average distance of Earth from the Sun is [tex]1.5\times10^{8}[/tex] kilometers.

Also, the approximate distance of Earth from the Moon is [tex]4\times10^{5}[/tex] kilometers.

For this, we will divide the average distance of Earth from the Sun by approximate distance of Earth from the Moon.

So,

[tex]\frac{1.5\times10^{8}}{4\times10^{5}}=375\ km[/tex]

Answer:

Null hypothesis:[tex]\mu_{1}-\mu_{0}\leq 0[/tex]

Alternative hypothesis:[tex]\mu_{1}-\mu_{2}>0[/tex]

[tex]SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705[/tex]

b) 2.70

[tex]t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850[/tex]

b. 1.85

[tex]p_v =P(Z>1.85)=0.032[/tex]

b. 0.03

a. We can reject H(0) in favor of H(a)

Step-by-step explanation:

Data given and notation

[tex]\bar X_{1}=164[/tex] represent the mean for the sample 1

[tex]\bar X_{2}=159[/tex] represent the mean for the sample 2

[tex]\sigma_{1}=12.5[/tex] represent the population standard deviation for the sample 1

[tex]s_{2}=9.25[/tex] represent the population standard deviation for the sample B2

[tex]n_{1}=35[/tex] sample size selected 1

[tex]n_{2}=30[/tex] sample size selected 2

[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

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

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the expenditure of households in City 1 is more than that of households in City 2, the system of hypothesis would be:

Null hypothesis:[tex]\mu_{1}-\mu_{0}\leq 0[/tex]

Alternative hypothesis:[tex]\mu_{1}-\mu_{2}>0[/tex]

We know the population deviations, so for this case is better apply a z test to compare means, and the statistic is given by:

[tex]z=\frac{(\bar X_{1}-\bar X_{2})-0}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}[/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".

Standard error

The standard error on this case is given by:

[tex]SE=\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}[/tex]

Replacing the values that we have we got:

[tex]SE=\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}=2.705[/tex]

b. 2.70

Calculate the statistic

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

[tex]t=\frac{(164-159)-0}{\sqrt{\frac{12.5^2}{35}+\frac{9.25^2}{30}}}=1.850[/tex]  

P-value

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

[tex]p_v =P(Z>1.85)=0.032[/tex]

b. 0.03

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis.

a. We can reject H(0) in favor of H(a)

Answer:

There is a 9.75% probability that exactly 2 of them have type O blood.

Step-by-step explanation:

For each children, there is only two possible outcomes. Either they have blood type O, or they do not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

In this problem we have that:

4 children, so [tex]n = 4[/tex]

Probability 0.15 of having blood type O, so [tex]p = 0.15[/tex]

If these parents have 4 children, whatthe probability that exactly 2 of them have type O blood?

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 2) = C_{4,2}.(0.15)^{2}.(0.85)^{2} = 0.0975[/tex]

There is a 9.75% probability that exactly 2 of them have type O blood.

Answer:

The tangent line equations are:

- The one that has the smaller slope [tex]\bold{y =3}[/tex]

- The one that has the larger slope [tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]

Step-by-step explanation:

In order to find the tangent lines we need to find first the first derivative since the first derivative evaluated at a point give us the slope of the tangent line.

Working with implicit differentiation.

In order to find the derivative we can work with implicit differentiation and we get

[tex]2x+18y \cfrac{dy}{dx}= 0[/tex]

Notice that only when we are finding the derivative of an expression that has y we multiply by dy/dx due chain rule.

Solving for the first derivative we get

[tex]18y\cfrac{dy}{dx}=-2x[/tex]

[tex]\cfrac{dy}{dx}=-\cfrac{x}{9y}[/tex]

So the equation of the first derivative at the point (x,y) give us the first equation for the slope.

[tex]m=-\cfrac{x}{9y}[/tex]

Slope of a line using definition.

We can also find the slope of the line that will pass the point (x,y) trough  the point (27,3) using the definition:

[tex]m = \cfrac{y_2-y_1}{x_2-x_1}[/tex]

Replacing the point we get

[tex]m=\cfrac{y-3}{x-27}[/tex]

That is another equation for the slope.

Finding the value of the slopes.

We can now set both slope equations equal to each other to find the point (x,y) where they intersect and the value of the slopes.

[tex]-\cfrac{x}{9y}= \cfrac{y-3}{x-27}[/tex]

We can work with cross multiplication to get

[tex]-x(x-27)=9y(y-3)[/tex]

And we can distribute and simplify

[tex]-x^2+27x=9y^2-27y[/tex]

Moving everything to the right side

[tex]0=x^2-27x+9y^2-27y\\\\0=x^2+9y^2-27x-27y[/tex]

At this point we can use the ellipse equation so we can replace [tex]x^2+9y^2[/tex] with 81, that will give us

[tex]0=81-27x-27y[/tex]

Then we can divide by 27 and solve for y.

[tex]0=3-x-y\\y= 3-x[/tex]

At this point we can replace on the ellipse equation.

[tex]x^2+9(3-x)^2=81[/tex]

And we can distribute and simplify.

[tex]x^2+9(9-6x+x^2)=81\\x^2+81-54x+9x^2=81\\x^2-54x+9x^2=0\\10x^2-54x=0\\[/tex]

So then we can factor and solve for x.

[tex]2x(5x-27)=0[/tex]

Setting each factor and solve for x we get

[tex]x= 0, \cfrac{27}{5}[/tex]

At x = 0, we can find the y value.

[tex]y= 3-x\\y= 3-0\\y= 3[/tex]

And the slope

[tex]m = -\cfrac{x}{9y}\\m = -\cfrac{0}{9y}\\m = 0[/tex]

So the line equation is

[tex]y-3=0(x-0)\\y=3[/tex]

For the second point x = 27/5 we have:

[tex]y= 3-\cfrac{27}{5}\\y=-\cfrac{12}{5}[/tex]

So slope is:

[tex]m = -\cfrac{\cfrac{27}{5}}{9\left(-\cfrac{12}{5}\right)}[/tex]

[tex]m = \cfrac 14[/tex]

So the line equation is

[tex]y-\left(-\cfrac{12}{5}\right)= \cfrac 14 \left(x -\cfrac{27}{5}\right)\\y = \cfrac 14 x - \cfrac{15}{4}[/tex]

Thus the equations of tangent lines are:

[tex]\bold{y =3}[/tex]

and

[tex]\bold{y = \cfrac 14 x - \cfrac{15}{4}}[/tex]

To find the equations of the tangent lines to the ellipse that passes through a specific point, differentiate the equation of the ellipse to get a formula for the slope at any point on the ellipse. Set this equal to the slope of the line from the point on the ellipse to the specific point, and solve. The resultant solutions give the points of tangency, and the lines through these points and the specific point are the desired tangent lines.

The subject of the question is about finding the equations of the tangent lines to the ellipse x^2 + 9y^2 = 81 that pass through a specific point (27, 3). We first differentiate the equation of the ellipse to get a general formula for the slope of the tangent line at any point on the ellipse. The differentiation of x^2 + 9y^2 = 81 with respect to x gives 2x + 18yy' = 0, so y' = -x/ (9y). Now, we put the slopes y' equal to the slopes of line from (x, y) to (27, 3), and solve the resulting system of equations. The solutions give the points of tangency, and the lines through these points and (27, 3) are the desired tangent lines. The exact computations involve solving a quadratic equation and it'll give two slopes for the tangent line that passes through the point (27,3).

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Answer:

The three statements are true.

Step-by-step explanation:

The question is incomplete:

A veterinarian has 70 clients who own cats, dogs, or both. Of these clients, 36 own cats, including 20 clients who own both cats and dogs. Which of the following statements must be true? Indicate all such statements.

A. There are 54 clients who own dogs.

B. There are 34 clients who own dogs but not cats.

C. There are 16 clients who own cats but not dogs.

A. There are 54 clients who own dogs.

TRUE. Of the 70 clients, only 36 own cats. There are left 34 clients that own only dogs. If we add the 20 clients that own both cats and dogs, we have 34+20=54 clients who own dogs.

B. There are 34 clients who own dogs but not cats.

TRUE. Of the 70 clients, only 36 own cats. Then, there are left 70-36=34 clients that own only dogs.

C. There are 16 clients who own cats but not dogs.

TRUE. Out of the 36 clients that own cats (only of with dogs), there are 20 that own both. Therefore, there are 36-20=16 clients that own only cats.

One way to do this is to recall that for [tex]|x|<1[/tex], we have

[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]

so that

[tex]\displaystyle\frac x{10x^2+1}=\frac x{1-(-10x^2)}=x\sum_{n=0}^\infty(-10x^2)^n=\sum_{n=0}^\infty(-10)^nx^{2n+1}[/tex]

(which seems to match the first option) so long as [tex]|-10x^2|=10x^2<1[/tex], or [tex]-\frac1{\sqrt{10}}<x<\frac1{\sqrt{10}}[/tex], which is the interval of convergence.

Answer:

i don't know sorry hope this helps

Step-by-step explanation:

Answer:

There is a 99.44% probability that the squad will have at most 2 calls in an hour.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given time interval.

In this problem, we have that:

2.83 rescues every eight hours.

What is the probability that the squad will have at most 2 calls in an hour?

This is

[tex]P = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

We have 2.83 rescues every 8 hours. So for an hour, we have [tex]\mu = \frac{2.83}{8} = 0.354[/tex]

So

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-0.354}*(0.354)^{0}}{(0)!} = 0.7019[/tex]

[tex]P(X = 1) = \frac{e^{-0.354}*(0.354)^{1}}{(1)!} = 0.2485[/tex]

[tex]P(X = 2) = \frac{e^{-0.354}*(0.354)^{2}}{(2)!} = 0.0440[/tex]

So

[tex]P = P(X = 0) + P(X = 1) + P(X = 2) = 0.7019 + 0.2485 + 0.0440 = 0.9944[/tex]

There is a 99.44% probability that the squad will have at most 2 calls in an hour.

Final answer:

The probability that the rescue squad will receive at most 2 calls in an hour is 0.9951

when the average is 2.83 calls every eight hours.

Explanation:

The number of rescue calls that a rescue squad receives is given to be a Poisson distribution with an average of 2.83 rescues every eight hours. To calculate the probability of receiving at most 2 calls in one hour, we first find the average number of rescues per hour by dividing the given average by eight, since there are eight hours in the period mentioned. This gives us an average rate ([tex]λ[/tex]) of 2.83 rescues / 8 hours = 0.35375 rescues per hour. The Poisson probability formula is:

[tex]P(X=k) = (e^{-λ} * λ^k) / k![/tex]

To find the probability of at most 2 calls, we sum the probabilities of 0, 1, and 2 calls:

[tex]P(X=0) = e^{-0.35375} * (0.353750)^0 / 0! = 0.7026 \\\\P(X=1) = e^{-0.35375} * (0.353751)^1 / 1! = 0.2485 \\\\P(X=2) = e^{-0.35375} * (0.353752)^2 / 2! = 0.0440[/tex]

Thus, the desired probability is the sum of these probabilities:

P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2) ≈ 0.7026 + 0.2485 + 0.0440 ≈ 0.9951

After rounding to four decimal places we have that the probability that the squad will have at most 2 calls in an hour is 0.9951.

Answer:

The amount fertilizer A the farmer should use while still meeting the minimum requirement is approximately seventy-six 50-lb bags which would cost $6080.

The amount of fertilizer B the farmer should use while still meeting the minimum requirement is approximately seventy-one 50- lb bags which would cost $2120

Step-by-step explanation:

Fertilizer A

Let x represent the number of 50- lb bags needed for the minimum requirement

Cost of I 50-lb bag = $80

Cost of x 50-lb bag = $80x

1 50-lb bag contains 14lb nutrient ( 8lb Nitrogen+ 2lb Phosphorus + 4lb Potassium)

x 50-lb bag contains 1060lb nutrient (440lb N + 260lb P + 360lb K) which is the minimum requirement

x = 1060lb ÷ 14lb = 75.71

x = 76 ( to the nearest whole number)

76 50-lb bags would cost 76×$80 = $6080

Fertilizer B

Cost of a 50- lb bag is $30

Let y be the number of bags required

Cost of y bag is $30y

y = 1060÷ 15= 71

Cost = 71× $30 = $2130

Answer:

0 fertilizer A

88 fertilizer B

Step-by-step explanation:

Shown in the attachment

Answer:

Option A.

Step-by-step explanation:

The given probability table is

Number of cars X: 0       1         2        3        4        5      6      7      8

Probability :0.087 0.323 0.363 0.144 0.053 0.019 0.007 0.002 0.001

It is given that a housing company builds houses with two car garages.

We need to find the percent of households have more cars than the garage can hold.

[tex]P(X>2)=1-P(X\leq 2)[/tex]

[tex]P(X>2)=1-[P(X=0)+P(X=1)+P(X=2)][/tex]

Substitute the probability values from the given table.

[tex]P(X>2)=1-[0.087+0.323+0.363][/tex]

[tex]P(X>2)=1-0.773[/tex]

[tex]P(X>2)=0.227[/tex]

It means 22.7% of households have more cars than the garage can hold.

Therefore, the correct option is A.

The percent of households that have more cars (including SUVs and light trucks) than a two-car garage can accommodate is approximately 22.7%.

To answer the student's question, let's examine the probability model provided. The random variable X represents the number of cars, including SUVs and light trucks, in a randomly selected American household.

If a housing company builds houses with two-car garages, we want to determine what percent of households have more cars than the garage can hold. To find this, we would add up the probabilities for the households that own more than 2 cars.

The probabilities for owning 3, 4, 5, 6, 7 or 8 cars are 0.144, 0.053, 0.019, 0.007, 0.002, and 0.001, respectively. Adding all these probabilities gives us:

0.144 + 0.053 + 0.019 + 0.007 + 0.002 + 0.001 = 0.226

To convert this to a percentage, we multiply by 100, which gives us 22.6%. Therefore, approximately 22.7% of households have more cars than the garage can hold, which corresponds to answer choice A.

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Answer:

[tex]E(X)= n\frac{M}{N}=10 \frac{5}{25}=2[/tex]

[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1[/tex]

Step-by-step explanation:

Previous concepts

The hypergeometric distribution is a discrete probability distribution that its useful when we have more than two distinguishable groups in a sample and the probability mass function is given by:

[tex]P(X=k)= \frac{(MCk)(N-M C n-k)}{NCn}[/tex]  

Where N is the population size, M is the number of success states in the population, n is the number of draws, k is the number of observed successes  

The expected value and variance for this distribution are given by:

[tex]E(X)= n\frac{M}{N}[/tex]

[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}[/tex]

What is the distribution of X?  

For this case the random variable X follows a hypergometric distribution.

Compute the values for E(X) and Var(X)  

For this case n=10, M=5, N=25, so then we can replace into the formulas like this:

[tex]E(X)= n\frac{M}{N}=10 \frac{5}{25}=2[/tex]

[tex]Var(X)=n \frac{M}{N}\frac{N-M}{N}\frac{N-n}{N-1}=10\frac{5}{25}\frac{25-5}{25}\frac{25-10}{25-1}=1[/tex]

What is the probability that none of the animals in the second sample are tagged?  

So for this case we want this probability:

[tex]P(X=0)= \frac{(5C0)(25-5 C 10-0)}{25C10}=\frac{1*184756}{3268760}=0.0565[/tex]  

What is the probability that all of the animals in the second sample are tagged?  

So for this case we want this probability:

[tex]P(X=5)= \frac{(5C5)(25-5 C 10-5)}{25C10}=\frac{1*15504}{3268760}=0.00474[/tex]  

Answer:

[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/tex]  

[tex]p_v =2*P(z>1.897)=0.0578[/tex]  

If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we see that [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 heads in the Euro coins is not significantly different from 0.5.  

Step-by-step explanation:

1) Data given and notation

n=250 represent the random sample taken

X=140 represent the number of heads obtained

[tex]\hat p=\frac{140}{250}=0.56[/tex] estimated proportion of heads

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha[/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 that one-Euro coins are biased, so the correct system of hypothesis are:  

Null hypothesis:[tex]p=0.5[/tex]  

Alternative hypothesis:[tex]p \neq 0.5[/tex]  

When we conduct a proportion test we need to use the z statisitc, 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].

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

[tex]np_o =250*0.5=125>10[/tex]

[tex]n(1-p_o)=250*(1-0.5)=125>10[/tex]

3) Calculate the statistic  

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

[tex]z=\frac{0.56 -0.5}{\sqrt{\frac{0.5(1-0.5)}{250}}}=1.897[/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>1.897)=0.0578[/tex]  

If we compare the p value obtained and the significance level assumed [tex]\alpha=0.05[/tex] we see that [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 heads in the Euro coins is not significantly different from 0.5.  

Answer:

0.3 or 30%

Step-by-step explanation:

Since no innocent student will ever be caught, the probability that a student sends an email and/or text message during a lecture AND gets caught is given by the product of the probability of a student sending a message (60%) by the probability of the professor catching them (50%) :

[tex]P = 0.5*0.6 = 0.3[/tex]

The probability is 0.3 or 30%.

Answer:

We conclude that the Pennsylvania school district have an IQ higher than the average of 101.5

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 101.5

Sample mean, [tex]\bar{x}[/tex] = 106.4

Sample size, n = 30

Alpha, α = 0.05

Population standard deviation, σ = 15

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 101.5\\H_A: \mu > 101.5[/tex]

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

Formula:

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

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{106.4 - 101.5}{\frac{15}{\sqrt{30}} } = 1.789[/tex]

Now, [tex]z_{critical} \text{ at 0.05 level of significance } = 1.96[/tex]

Since,  

[tex]z_{stat} > z_{critical}[/tex]

We reject the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that the Pennsylvania school district have an IQ higher than the average of 101.5