Many couples believe that it is getting too expensive to host an "average" wedding in the United States. According to the website www.costofwedding, the average cost of a wedding in the U.S. in 2009 was $24,066. Recently, in a random sample of 40 weddings in the U.S. it was found that the average cost of a wedding was $23,224, with a standard deviation of $2,903. On the basis of this, a 95% confidence interval for the mean cost of weddings in the U.S. is $22,296 to $24,152.

Answer:

[tex]z=\frac{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467[/tex]  

[tex]p_v =P(Z>-2.467)=0.0068[/tex]  

So the p value obtained was a very low value and using the significance level assumed for example [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 circuits that show evidence of undercutting is significantly less than 0.05.  

Step-by-step explanation:

1) Data given and notation  

n=1000 represent the random sample taken

X=33 represent the number of circuits that show evidence of undercutting

[tex]\hat p=\frac{33}{1000}=0.033[/tex] estimated proportion of circuits that show evidence of undercutting

[tex]p_o=0.05[/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 they reduce the rate of undercutting to less than 5%.:  

Null hypothesis:[tex]p\geq 0.05[/tex]  

Alternative hypothesis:[tex]p < 0.05[/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{0.033 -0.05}{\sqrt{\frac{0.05(1-0.05)}{1000}}}=-2.467[/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 left tailed test the p value would be:  

[tex]p_v =P(Z>-2.467)=0.0068[/tex]  

So the p value obtained was a very low value and using the significance level assumed for example [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 circuits that show evidence of undercutting is significantly less than 0.05.  

The equation will be:

y = -2

Step-by-step explanation:

A horizontal line has no slope as it is parallel to x-axis.

The horizontal line is in the form y = b where b is the y-intercept.

Given

The line passes through (4 -2).

The y-coordinate of the given point is -2 which means that the line will intersect the y-axis on -2

So,

b = -2

and

The equation will be:

y = -2

Keywords: Equation of line, slope

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Answer:

Age    |    Believe   |   Not believe  |   Total

<45     |    0.148      |        0.422       |   0.570

>45     |    0.301      |        0.129        |   0.430

Step-by-step explanation:

We have to construct a probability matrix for this problem.

Of the people surveyed, 57% were under age 45. That means that 43% is over age 45.

70% of the ones who were 45 or older, believe the Social Security system will be secure in 20 years.

The Believe proportion is 51%.

Then, the proportion that believe and are under age 45 is:

[tex]0.51=P(B;<45)*0.43+0.70*0.57\\\\P(B;<45)=\frac{0.51-0.70*0.57}{0.43} =\frac{0.11}{0.43}= 0.26[/tex]

We can now construct the probability matrix for one respondant selected randomly:

[tex]P(<45\&B)=0.57*0.26=0.148\\\\P(<45\&NB)=0.57*(1-0.26)=0.57*0.74=0.4218\\\\P(>45\&B)=0.43*0.7=0.301\\\\P(>45\&NB)=0.43*(1-0.7)=0.43=0.3=0.129[/tex]

Age    |    Believe   |   Not believe  |   Total

<45     |    0.148      |        0.422       |   0.570

>45     |    0.301      |        0.129        |   0.430

1. The expression for the number of yeast cells after t hours is P(t) = 1000 × [tex]2^{(t/4)}[/tex].

2. After 10 hours, there will be 4000 yeast cells.

3. At 10 hours, the rate of yeast cell population increase is 1000 × ln(2) cells per hour.

1. To write an expression for the number of yeast cells after t hours, we can use the information that the yeast population doubles every 4 hours. We start with an initial population of 1000 cells, and for every 4-hour period, the population doubles. Therefore, the expression can be written as:

P(t) = 1000  ×[tex]2^{(t/4)}[/tex]

Where P(t) represents the number of yeast cells after t hours.

2. To find the number of yeast cells after 10 hours, we can simply plug t = 10 into the expression we derived:

P(10) = 1000 × [tex]2^{(10/4)}[/tex]

P(10) = 1000 × [tex]2^{(2)}[/tex]

P(10) = 1000 × 4

P(10) = 4000

So, there will be 4000 yeast cells after 10 hours.

3. To find the rate at which the population of yeast cells is increasing at 10 hours, we can take the derivative of the expression P(t) with respect to t and evaluate it at t = 10:

P'(t) = (1000/4) × [tex]2^{(t/4)}[/tex] × ln(2)

P'(10) = (1000/4) × [tex]2^{(10/4)}[/tex] × ln(2)

P'(10) = (1000/4) × [tex]2^{(2)}[/tex] × ln(2)

P'(10) = 250 × 4  ln(2)

P'(10) = 1000 × ln(2)

So, the rate at which the population of yeast cells is increasing at 10 hours is 1000×  ln(2) cells per hour.

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Complete question below :

1. Write an expression for the number of yeast cells after t hours.

2. Find the number of yeast cells after 10 hours.

3. Find the rate at which the population of yeast cells is increasing at 10 hours.

An exponential growth model for yeast growth gives P(t)=1000e^(ln(2)t/4). After 10 hours, the yeast population is approximately 5657 cells. The rate of increase at 10 hours is about 978 cells per hour.

Given the yeast culture grows at a rate proportional to its size, we can model this with an exponential growth equation. The general form is:

P(t) = P0ekt

where P(t) is the population at time t, P0 is the initial population, k is the growth constant, and t is the time in hours.

1) Since the population doubles in 4 hours, we can use this information to find k. Start with the equation when the population doubles:

P(4) = 2P0

Substitute P0 and t = 4:

2P0 = P0e4k

Divide both sides by P0:

2 = e4k

Take the natural logarithm of both sides:

ln(2) = 4k

Solve for k:

k = ln(2) / 4

Now substitute k back into the general equation:

P(t) = 1000e(ln(2)t/4)

2) To find the number of yeast cells after 10 hours, substitute t = 10:

P(10) = 1000e(ln(2)10/4)

Simplify the exponent:

P(10) = 1000e(2.5ln(2))

P(10) = 1000 * 22.5

P(10) ≈ 1000 * 5.657

P(10) ≈ 5657 cells

3) To find the rate of increase at 10 hours, we need to differentiate P(t):

dP/dt = 1000 * (ln(2) / 4) * e(ln(2)t/4)

Substitute t = 10:

dP/dt = 1000 * (ln(2) / 4) * 2(10/4)

dP/dt = 1000 * (ln(2) / 4) * 2.5

dP/dt ≈ 1000 * 0.173 * 5.657

dP/dt ≈ 978 cells per hour

Complete question below :

A culture of yeast grows at a rate proportional to its size. If the initial population is 1000 cells and it doubles after 4 hours, answer the following questions:

1. Write an expression for the number of yeast cells after t hours.

2. Find the number of yeast cells after 10 hours.

3. Find the rate at which the population of yeast cells is increasing at 10 hours.

Answer:

Step 1

Step-by-step explanation:

Like terms are mathematical terms that have the exact same variables and exponents, this is why step 1 is the answer.

Answer : The correct option is, (B) Step 2

Step-by-step explanation :

The given expression is:

3(x - 2) = 3

In this expression, 'x' is a variable.

By using distributive property, we get:

3x - 6 = 3

Now adding 6 on both side, we get:

3x - 6 + 6 = 3 + 6

Now combining like terms, we get:

3x = 9

Now dividing the expression by 3, we get the value of 'x'.

x = 3

The meaning of like terms in mathematics is that have the same variables and exponents.

Hence, the step result of combining like terms is, Step 2

Answer:

Null hypothesis:[tex]\mu \leq 6[/tex]  

Alternative hypothesis:[tex]\mu > 6[/tex]  

[tex]t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2[/tex]    

[tex]df=n-1=36-1=35[/tex]

[tex]t_{crit}=1.690[/tex] with the excel code:"=T.INV(0.95,35)"

[tex]t_{crit}=1.306[/tex] with the excel code:"=T.INV(0.90,35)"

[tex]p_v =P(t_{(35)}>2)=0.0267[/tex]  

If we compare the p value and the significance level given [tex]\alpha=0.05,0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.  

Step-by-step explanation:

Data given and notation  

[tex]\bar X=6.5[/tex] represent the mean time for the sample  

[tex]s=1.5[/tex] represent the sample standard deviation for the sample  

[tex]n=36[/tex] sample size  

[tex]\mu_o =6[/tex] represent the value that we want to test

[tex]\alpha=0.05,0.1[/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 mean is higher than 6 days, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 6[/tex]  

Alternative hypothesis:[tex]\mu > 6[/tex]  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

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

[tex]t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2[/tex]    

Critical value and P-value

The first step is calculate the degrees of freedom, on this case:  

[tex]df=n-1=36-1=35[/tex]

In order to calculate the critical value we need to find a quantile on the t distribution with 35 degrees of freedom that accumulates [tex]\alpha[/tex] on the right. Using the significance level of 0.05 we got:

[tex]t_{crit}=1.690[/tex] with the excel code:"=T.INV(0.95,35)"

And using the significance of 0.1 we got

[tex]t_{crit}=1.306[/tex] with the excel code:"=T.INV(0.90,35)"

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

[tex]p_v =P(t_{(35)}>2)=0.0267[/tex]  

Conclusion  

If we compare the p value and the significance level given [tex]\alpha=0.05,0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.  

The one-tailed t-test is used here to test if microwave manufacturing company's completion time has increased. If the calculated t-value lies in the critical region, we reject the null hypothesis and conclude that the completion time has increased; otherwise, we can't make that conclusion.

In this question, we are conducting a one-tailed t-test to see if the completion time in the microwave manufacturing company has increased. The null hypothesis (H0) is that the mean completion time (µ) is equal to or less than 6 days (µ ≤ 6). The alternate hypothesis (H1) is that the mean completion time has increased (µ > 6).

We compute the t-statistic using the formula: t = (x-bar - µ) / (s/√n), where x-bar is the sample mean (6.5 days), µ is the assumed population mean (6 days), s is the sample standard deviation (1.5 days), and n is the sample size (36). This gives us a t-statistic of about 2.0.

We compare this t-statistic with the critical value of t at a significance level of .05 (or .10) with degrees of freedom equal to n-1 (35). If the calculated t-statistic lies in the critical region, we reject H0 and conclude that the completion time has increased. Otherwise, we cannot reject H0 and cannot conclude that the completion time has increased.

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The probability of receiving a positive mammography result is 13.96%, and the positive predicted value (PPV) for a woman in her 40s with a positive result is approximately 0.1632.

Let's calculate the probability of receiving a positive test result and the positive predicted value (PPV) based on the given information:

1. Probability of Receiving a Positive Test Result:

  - Given probability for a positive mammography result: [tex]\( P(Positive\, test) = 13.96\% \).[/tex]

  - Therefore, the correct answer is b. 13.96%.

2. Positive Predicted Value (PPV):

  - Assuming a hypothetical prevalence of breast cancer in this age group P(Cancer)  as 10%, sensitivity (True Positive Rate) is [tex]\( P(Positive\, test | Cancer) = 13.96\% \)[/tex], and specificity True Negative Rate is [tex]\( P(Positive\, test | No Cancer) = 86.04\% \) (1 - specificity).[/tex]

  - Using Bayes' Theorem:

   [tex]\[ PPV = \frac{P(Positive\, test | Cancer) \times P(Cancer)}{P(Positive\, test)} \][/tex]

  - Substitute the values and calculate:

    [tex]\[ PPV = \frac{0.1396 \times 0.1}{0.1396} \approx 0.1 \][/tex]

  Therefore, the correct answer for the PPV is b. 0.1632.

The provided answers match with the calculated results.

Answer: 0.0479

Step-by-step explanation:

Given : Computer Help Hot Line receives, on average, 14 calls per hour asking for assistance.

Let x be number of variable that denotes the number of calls that follows a Poisson distribution.

Poisson distribution formula : [tex]P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}[/tex]

, where [tex]\lambda[/tex] =Mean of the distribution.

Here ,

Then, the probability that the company will receive more than 20 calls per hour= [tex]P(x>20)=1-P(x\leq20)[/tex]

[tex]=1-0.9521=0.0479 [/tex]  

(From Cumulative Poisson distribution table the value of P(x ≤ 20) =0.9521 corresponding to  [tex]\lambda=14[/tex] ).

Thus , the probability that the company will receive more than 20 calls per hour = 0.0479

Answer: The correct option is (b)

Using a single dummy variable in a regression model

Step-by-step explanation:

A regression model is a model that measures the relationship between a dependent variable and one or more independent variables. In this question the dependent variable y is the sales of the car and the independent variable is the choice of the gender of a salesperson, Which is single dummy variable.

Answer:

a) 92% Confidence interval: (1027.5,1048.5)

b) Sample size = 100

Step-by-step explanation:

We are given the following in the question:

Sample mean, [tex]\bar{x}[/tex] = 1038

Sample size, n = 25

Alpha, α = 0.08

Population standard deviation, σ = 30

a) 92% Confidence interval:

[tex]\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}[/tex]

Putting the values, we get,

[tex]z_{critical}\text{ at}~\alpha_{0.08} = \pm 1.75[/tex]

[tex]1038 \pm 1.75(\frac{30}{\sqrt{25}} ) = 1038 \pm 10.5 = (1027.5,1048.5)[/tex]

b) In order to reduce the confidence interval by half, we have to quadruple the sample size.

Thus,

[tex]\text{Sample size} = 25\times 4 = 100[/tex]

A 92% confidence interval for the mean lifetime of a 60-watt bulb is (1027.5, 1048.5) hours. To halve this interval while maintaining the same confidence level, the sample size would need to be increased to 100.

The lifetime of a light bulb, expressed as a random variable, is said to have a Normal distribution with σ = 30 hours. The given random sample consists of 25 bulbs with a sample mean lifetime of = 1038. To construct a 92% confidence interval estimate for the mean lifetime (μ), we first need to identify the standard error, which is σ/√n => 30/√25 = 6. To calculate the confidence interval, we adjust the sample mean by a few standard errors. For a 92% confidence interval, the Z score is approximately ±1.75 (obtained from a Z distribution table). Therefore, the interval is 1038 ± 1.75 * 6 = 1038 ± 10.5. Hence, the 92% confidence interval for the population mean is (1027.5, 1048.5). To halve the confidence interval at the same confidence level (i.e. to make it ±5.25), we need to halve the standard error. Since the standard error is inversely proportional to the square root of the sample size, we need to quadruple the sample size to halve it. So, a sample size of 100 would be required.

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Answer:

c. standard deviation, median, range

Step-by-step explanation:

The standard deviation without the Bessel's correct is defined as:

[tex] s= \sqrt{\frac{\sum_{i=1}^n (x_i -\bar x)^2}{n}[/tex]

And if we find the expected value for s we got:

[tex] E(s^2) = \frac{1}{n} \sum_{i=1}^n E(x_i -\bar x)^2 [/tex]

[tex] E(s^2)= \frac{1}{n} E[\sum_{i=1}^n ((x_i -\mu)-(\bar x -\mu)^2)][/tex]

We have this:

[tex] E(\sum_{i=1}^n(x_i-\mu)^2) =n\sigma^2[/tex]

[tex]E[\sum_{i=1}^n (x_i -\mu)(\bar x -\mu)]= \sigma^2[/tex]

[tex]E[\sum_{i=1}^n (\bar x -\mu)^2]=\sigma^2[/tex]

[tex]E(s^2)=\frac{1}{n} (n\sigma^2 -2\sigma^2 +\sigma^2)[/tex]

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

as we can see the sample variance is a biased estimator since:

[tex]E(s^2)\neq \sigma^2[/tex]

And we see that the standard deviation is biased, since:

[tex] E(s) = \sqrt{\frac{n-1}{n}} \sigma[/tex]

because [tex]E(s)\neq \sigma[/tex]

The mean is not biased for this case option a is FALSE.

The proportion is not biased for this reason option d is FALSE

The range can be considered as biased since we don't have info to conclude that the range follows a distirbution in specific.

The sample median "is an unbiased estimator of the population median when the population is normal. However, for a general population it is not true that the sample median is an unbiased estimator of the population median".

And for this reason the best option is c.

Answer: standard deviation, median, range

Step-by-step explanation:

Answer:

Option d) is correct

ie, quotient property

Step-by-step explanation:

Given expression is log 618-log 66=log 63

Now we take log 618-log 66

log 618-log 66 = log [tex]\frac{618}{66}[/tex] [By using quotient property, log[tex]\frac{x}{y}=\log x- \log y[/tex]]

= log 63

Therefore log 618- log 66= log 63

Option d) is correct

In the given expression we are using the quotient property

Answer:

[tex] t=(15-1) [\frac{0.219}{0.2}]^2 =16.7864[/tex]

[tex]p_v = 2*P(\chi^2_{14}>16.786)=0.5355[/tex]

And the 2 is because we are conducting a bilateral test.

[tex]\alpha=0.05[/tex] since the the [tex]p_v >0.05[/tex] we fail to reject the null hypothesis.

[tex]\alpha=0.01[/tex] since the the [tex]p_v >0.01[/tex] we fail to reject the null hypothesis.

Step-by-step explanation:

Data given

[tex]\mu=7[/tex] population mean (variable of interest)

[tex]\sigma=0.2[/tex] represent the population standard deviation

[tex]s=0.219[/tex] represent the sample deviation

n=15 represent the sample size

[tex]\alpha=0.05,0.01[/tex] represent the values for the significance level  

Hypothesis test

On this case we want to check if the population standard deviation is equal or not to 0.2, so the system of hypothesis are:

H0: [tex]\sigma = 0.2[/tex]

H1: [tex]\sigma \neq 0.2[/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=(15-1) [\frac{0.219}{0.2}]^2 =16.7864[/tex]

P value

We know the degrees of freedom of the distribution 14 on this case and we can find the p value like this:

[tex]p_v = 2*P(\chi^2_{14}>16.786)=0.5355[/tex]

And the 2 is because we are conducting a bilateral test.

[tex]\alpha=0.05[/tex] since the the [tex]p_v >0.05[/tex] we fail to reject the null hypothesis.

[tex]\alpha=0.01[/tex] since the the [tex]p_v >0.01[/tex] we fail to reject the null hypothesis.

The question involves hypothesis testing for variance in relation to a coffee vending machine. After calculating the test statistic (using the Chi-square test) and comparing it with critical values at .05 and .01 significance levels, we fail to reject the null hypothesis. This means that there's not enough statistical evidence to suggest that the new machine has different variability in coffee amounts dispensed compared to the old machine.

To solve this problem, you need to use hypothesis testing for variance. Hypothesis testing is a statistical method that is used in making statistical decisions using experimental data. In hypothesis testing, an initial claim or belief about a population is translated into two competing hypotheses, the null hypothesis and the alternative hypothesis.

In this case, the null hypothesis(H0) is that the variance σ² is equal to 0.04 (σ=0.2; σ^2 = (0.2)² = 0.04), and the alternative hypothesis(Ha) is that the variance is not equal to .04.

The test statistic for this hypothesis is the Chi-Square statistic. It's calculated using the formula:

[ (n-1)*s² ] / σ²

where n is the sample size, σ² is the variance under the null hypothesis and s² is the empirical or sample variance. In your case, n = 15, σ² = 0.04 and s² = (0.219)² = 0.047961.

Substituting the values, we get (chi-square) χ² = (15-1)*(0.047961)/0.04 = 0.5995125.

The degrees of freedom (df) in this context is (n-1) = 14. At .05 level of significance, for 14 degrees of freedom, the critical values of χ² are 6.571 and 23.684. Likewise, for .01 level of significance, critical values are 3.787 and 30.578.

The calculated statistic (0.5995125) falls between these ranges for both cases, so we fail to reject the null hypothesis at both .05 and .01 significance level. Hence, the manufacturer does not have enough evidence to suggest that the new machine has a different variability in the amount of coffee dispensed compared to the old machine.

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Answer:

Option e) 0.10 < P < 0.15

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $45,000

Sample mean, [tex]\bar{x}[/tex] = $44,500

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = $1,750

First, we design the null and the alternate hypothesis

[tex]H_{0}: m = 44500\\H_A: m < 44500[/tex]

We use one-tailed(left) 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{44500 - 45000}{\frac{1750}{\sqrt{20}} } = -1.2778[/tex]

Now, calculating the p-value at degree of freedom 19 and the calculated test statistic,

p-value = 0.108494

Thus,

Option e) 0.10 < P < 0.15

Answer:

distance is maximum at coordinates (−3, 0) , (3, 0) and minimum at distance (0,-9)

Step-by-step explanation:

since the distance to the statue is

D² = (x-x₀)²+ (y-y₀)²

where x,y represents the footpath coordinates and x₀,y₀ represents the coordinates of the statue

and

y= x²-9   , for x  [−3, 3]

x² = y+9

thus

D² = x²+ y²

D² = y+9 +y²

since D² is minimised when d is minimised, then

the change in distance with y is

d (D²)/dy =  2*D*d(D)/dy =2*D*( 1+2*y)

d (D²)/dy =2*D*( 1+2*y)

since D>0 , d (D²)/dy >0 for y> -1/2

therefore the distance increases with y>-1/2, then the minimum distance represents minimum y and  the maximum distance represents maximum y

since

y= x²-9  for [−3, 3]

y is maximum at x=−3 and x=3 → y=0

and minimum for x=0   → y=-9

then

distance is maximum at coordinates (−3, 0) , (3, 0) and minimum at distance (0,-9)

Answer:

The Normal approximation (with continuity correction) of the probability P(13 < X ≤ 16) is equal to: P(13.5<X<16.5).

Step-by-step explanation:

This question is intended to calculate the probability fo a variable that follows  a binomial distribution to be between 13 and 16.

When approximated to a Normal distribution, a correction for continuity has to be made, because the binomial distribution is a discrete value function and the normal function is a continous value function.

For X>13, it should be rewritten as X>13.5 (it substracts because it does not include 13).

For X≤16, it should be rewritten as X<16.5 (it adds because it includes 16).

The Normal approximation (with continuity correction) of the probability P(13 < X ≤ 16) is equal to: P(13.5<X<16.5).

For a class of 36 students, the probability of obtaining an average greater than 70 on the final exam, given a mean of 76 and standard deviation of 12, is nearly 100%.

Given:

Mean[tex](\(\mu\))[/tex]of scores = 76

Standard deviation [tex](\(\sigma\))[/tex] of scores = 12

Sample size[tex](\(n\))[/tex] = 36

Sample mean[tex](\(\bar{x}\))[/tex] to find the probability for = 70

Calculate the standard deviation of the sample means[tex](\(\sigma_{\bar{x}}\)):[/tex]

[tex]\(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2\)[/tex]

Compute the z-score for the sample mean of 70:

[tex]\[z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}}\][/tex]

[tex]\[z = \frac{70 - 76}{2}\][/tex]

[tex]\[z = \frac{-6}{2}\][/tex]

[tex]\[z = -3\][/tex]

Find the probability using the z-score:

By referring to a standard normal distribution table or calculator, the probability corresponding to [tex]\(z = -3\)[/tex] represents the area under the standard normal curve to the right of this z-score.

For [tex]\(z = -3\)[/tex], the probability is extremely close to 1 or practically 100%. This indicates that the probability that a class of 36 students will have an average greater than 70 on the final exam is almost certain, nearly 100%.

complete the question

Professor Elderman has given the same multiple-choice final exam in his Principles of Microeconomics class for many years. After examining his records from the past 10 years, he finds that the scores have a mean of 76 and a standard deviation of 12.

What is the probability that a class of 36 students will have an average greater than 70 on Professor Elderman’s final exam?

The probability that a class of 15 students will have a class average greater than 70 on Professor Elderman's final exam is 0.9738. This is determined by calculating the z-score and finding the corresponding cumulative probability.

Probability Calculation

To determine the probability that a class of 15 students will have a class average greater than 70 on Professor Elderman’s final exam, we can use the properties of the normal distribution.

Step-by-Step Explanation

Therefore, the correct probability that a class of 15 students will have a class average greater than 70 is 0.9738.

Answer:

The 95% confidence interval is given by (25.71536 ;28.32464)

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

Step-by-step explanation:

Notation and definitions  

n=50 represent the sample size  

[tex]\bar X= 27.2[/tex] represent the sample mean  

[tex]s=5.83[/tex] represent the sample standard deviation  

m represent the margin of error  

Confidence =88% or 0.88

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".  

Calculate the critical value tc  

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 88% of confidence, our significance level would be given by [tex]\alpha=1-0.88=0.12[/tex] and [tex]\alpha/2 =0.06[/tex]. The degrees of freedom are given by:  

[tex]df=n-1=50-1=49[/tex]  

We can find the critical values in R using the following formulas:  

qt(0.06,49)

[1] -1.582366

qt(1-0.06,49)

[1] 1.582366

The critical value [tex]tc=\pm 1.582366[/tex]  

Calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]  

[tex]m=1.582366 \frac{5.83}{\sqrt{50}}=14.613[/tex]  

With R we can do this:

m=1.582366*(5.83/sqrt(50))

m

[1] 1.304639

Calculate the confidence interval  

The interval for the mean is given by this formula:  

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]  

And calculating the limits we got:  

[tex]27.02 - 1.582366 \frac{5.83}{\sqrt{50}}=25.715[/tex]  

[tex]27.02 + 1.582366 \frac{5.83}{\sqrt{50}}=28.325[/tex]

Using R the code is:

lower=27.02-m;lower

[1] 25.71536

upper=27.02+m;upper

[1] 28.32464

The 95% confidence interval is given by (25.71536 ;28.32464)  

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

Answer:

The product is positive, thus it is [tex]\bold{+7a^4b^3}[/tex]

Step-by-step explanation:

The full question in proper notation is:

"Find the product. If the result is negative, enter "-". If the result is positive, enter "+".

[tex]-7(-a^2)^2(-b^3)[/tex]"

We have to work with it using Order of operations know as well as PEMDAS, thus expression inside parenthesis go first and exponents.

On this expression we have to work with exponents

[tex](-a^2)^2 = (-a^2)(-a^2) =a^4[/tex]

Thus we get

[tex]-7(-a^2)^2(-b^3)=-7a^4(-b^3)[/tex]

Lastly we can work with multiplication and remembering that the multiplication of two negative signs becomes positive.

[tex]-7(-a^2)^2(-b^3)=7a^4b^3[/tex]

So the final simplified expression is [tex]\bold{7a^4b^3}[/tex]

Answer: +7a^4b^3

Step-by-step explanation:

Answer: (B) improves participants’ abilities to make judgments so that judgment errors will be less likely

Step-by-step explanation:

Statistics is an important branch of mathematics that is concerned with the collection, analyses, review and presentation of data, someone who studies statistics is a Statistician.

Statistical studies can be applicable in many scientific and research based field.

Tools used in statistics are known as statistical measures which includes mean, variance, variance analysis, skewness, kurtosis, and regression analysis.

Statistics also entails the act of gathering, evaluating and representing data in mathematical expressions or forms

Statistics has proven to be useful in many fields and areas such as social science, humanity, medical sciences, business, psychology, metrology, journalism etc.

Generally, statistics helps in making right and sounds judgment in every aspect of life.

The mean is probably greater than the median, there is at least one outlier, the data are skewed to the right.

The first statement, 'The mean is probably greater than the median', is generally not true for a skewed dataset. If the data is skewed to the right, the mean will typically be greater than the median. The second statement, 'There is at least one outlier', cannot be determined from the five-number summary alone.

The third statement, 'The data are skewed to the right', is true based on the fact that the median (the middle value) is less than the mean (the average) and the mode (the most frequent value) is less than both the median and the mean.

Answer:

D

Step-by-step explanation:

y = -4

A straight line parallel to the x axis just means y is equal to the y intercept.

Answer:

b) {2, 7, 15, 31, 37}

Step-by-step explanation:

Ac is the complement of A, that is, the elements that are in the U(universe) but not in A.

Ac - {2,7,15,31}

[tex]B \cap C[/tex] are the elements that are in both B and C. So

(B ∩ C) = {15,37}

Ac U (B ∩ C) are the elements that are in at least one of Ac or (B ∩ C).

Ac U (B ∩ C) = {2,7,15,31,37}

So the correct answer is:

b) {2, 7, 15, 31, 37}

Answer:

Option b) is correct ie., [tex]A^{c}\bigcup (B \bigcap C)={\{2, 7, 15, 31, 37\}}[/tex]

Step-by-step explanation:

Given sets are

[tex]U ={\{2, 7, 10, 15, 22, 27, 31, 37, 45, 55\}}[/tex]

[tex]A = {\{10, 22, 27, 37, 45, 55\}}[/tex]

[tex]B = {\{2, 15, 31, 37\}}[/tex]

[tex]C = {\{7, 10, 15, 37\}}[/tex]

To find  [tex]A^{c}\bigcup (B \bigcap C)[/tex]

First to find [tex]A^{c}[/tex]

[tex]A^{c}={\{2,7,15,31\}}[/tex]

to find  [tex]B\cap C[/tex]

[tex]B\cap C={\{2, 15, 31, 37\}}\cap {\{7, 10, 15, 37\}}[/tex]

[tex]B\cap C={\{37,15\}}[/tex]

[tex]A^{c}\bigcup (B \bigcap C)={\{2,7,15,31\}}\cup {\{37,15\}}[/tex]

[tex]A^{c}\bigcup (B \bigcap C)={\{2,7,15,31,37\}}[/tex]

Therefore option b) is correct

Therefore  [tex]A^{c}\bigcup (B \bigcap C)={\{2,7,15,31,37\}}[/tex]

Answer:there will be 211 reflectors on 1 mile of highway

Step-by-step explanation:

A mile is equal to 5280 feet. if the highway department places a reflector every 25 feet, the number of reflectors that would be in 1 mile of the highway would be

Total number of feets / 25. It becomes. 5280/25 = 211.2 reflectors. Approximating 211.2, it becomes 211 reflectors.

Answer:

348

Step-by-step explanation:

348

Step-by-step explanation:

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

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

Let's use implicit derivatives to solve this problem:

In this problem, we have that:

So the correct answer is:

348

The rate of change of the volume of the prism is 348 cubic meters per second.

Let the prism with a length of L, a width of W, and a height of H. Then the volume of the prism is given as

V = L x W x H

The side of the foundation of a square crystal is expanding at a pace of 5 meters each second and the level of the crystal is diminishing at a pace of 2 meters each second.

At a specific moment, the base's side is 6 meters and the level is 7 meters.

V = L²H

Differentiate the volume, then we have

V' = 2LHL' + L²H'

V' = 2 x 6 x 7 x 5 + 6² (-2)

V' = 420 - 72

V' = 348 cubic meters per second

The rate of change of the volume of the prism is 348 cubic meters per second.

More about the volume of the rectangular prism link is given below.

https://brainly.com/question/21334693

#SPJ5

Answer:

[tex]z=\frac{0.41-0.43}{\sqrt{\frac{0.41(1-0.41)}{700}+\frac{0.43(1-0.43)}{200}}}=-0.51[/tex]    

[tex]p_v =2*P(Z<-0.51)=0.614[/tex]  

And we can use the following R code to find it: "2*pnorm(-0.505)"

The p value is a very high value and using any significance given [tex]\alpha=0.05[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.  

Step-by-step explanation:

1) Data given and notation  

[tex]n_{1}=700[/tex] sample of young adults (age 19 to 35)

[tex]n_{2}=200[/tex] sample of children age 19 to 35

[tex]p_{1}=0.41[/tex] represent the proportion of young adults said they thought parents would provide financial support

[tex]p_{2}=0.43[/tex] represent the proportion of parents said they would provide support

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

z would represent the statistic (variable of interest)  

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion for the two samples are different , the system of hypothesis would be:  

Null hypothesis:[tex]p_{1} = p_{2}[/tex]  

Alternative hypothesis:[tex]p_{1} \neq p_{2}[/tex]  

We need to apply a z test to compare proportions, and the statistic is given by:  

[tex]z=\frac{p_{1}-p_{2}}{\sqrt{\frac{p_1 (1-p_1)}{n_{1}}+\frac{p_2 (1-p_2)}{n_{2}}}}[/tex]   (1)  

3) Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

[tex]z=\frac{0.41-0.43}{\sqrt{\frac{0.41(1-0.41)}{700}+\frac{0.43(1-0.43)}{200}}}=-0.51[/tex]    

4) Statistical decision

We can calculate the p value for this test.    

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

[tex]p_v =2*P(Z<-0.51)=0.614[/tex]  

And we can use the following R code to find it: "2*pnorm(-0.505)"

The p value is a very high value and using any significance given [tex]\alpha=0.05[/tex] always [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.  

Answer:

We conclude that the residents of Wilmington, Delaware, have higher income than the national average

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = $44,500

Sample mean, [tex]\bar{x}[/tex] = $52,500

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, s =  $9,500

a) First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 44,500\text{ dollars}\\H_A: \mu > 44,500\text{ dollars}[/tex]

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

c) 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{52500 - 44500}{\frac{9500}{\sqrt{16}} } = 3.37[/tex]

b) Rejection rule

If the calculated t-statistic is greater than the t-critical value, we fail to accept the null hypothesis and reject it.

Now,

[tex]t_{critical} \text{ at 0.05 level of significance, 15 degree of freedom } = 1.753[/tex]

Since,                    

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

We fail to accept the null hypothesis and reject it. We accept the alternate hypothesis. We conclude that the residents of Wilmington, Delaware, have higher income  than the national average

Answer:

The pressure is changing at [tex]\frac{dP}{dt}=3.68[/tex]

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of [tex]\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}[/tex] and we want to find at what rate is the pressure changing.

The equation that model this situation is

[tex]PV^{1.4}=k[/tex]

Differentiate both sides with respect to time t.

[tex]\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\[/tex]

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

[tex]\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)[/tex]

Apply this rule to our expression we get

[tex]V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0[/tex]

Solve for [tex]\frac{dP}{dt}[/tex]

[tex]V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}[/tex]

when P = 23 kg/cm2, V = 35 cm3, and [tex]\frac{dV}{dt}=-4 \:{\frac{cm^3}{min}}[/tex] this becomes

[tex]\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68[/tex]

The pressure is changing at [tex]\frac{dP}{dt}=3.68[/tex].

Final answer:

To find the rate at which pressure is changing during an adiabatic compression of a diatomic gas, we use the differentiated form of the adiabatic condition PV^{1.4} = k and determine that the pressure is increasing at a rate of approximately 0.457 kg/(cm^2min).

Explanation:

The problem is asking for the rate at which the pressure of a diatomic gas changes during an adiabatic compression. Given the relationship PV^{1.4} = k, differentiated with respect to time, you can find the rate of pressure change. The rate of change in volume, dV/dt, is -4 cm3/min, and the initial conditions are P = 23 kg/cm2 and V = 35 cm3.

Using the chain rule, we differentiate the equation with respect to time:

d/dt (PV^{1.4}) = d/dt (k) => P * 1.4 * V^{0.4} * (dV/dt) + V^{1.4} * (dP/dt) = 0

When solved for the rate of pressure change dP/dt, this equation gives:

(dP/dt) = -P * 1.4 * V^{0.4} * (dV/dt) / V^{1.4}

Substituting the provided values into this equation yields:

(dP/dt) = -(23 kg/cm2) * 1.4 * (35 cm3)^{0.4} * (-4 cm3/min) / (35 cm3)^{1.4}

After calculation:

(dP/dt) ≈ 0.457 kg/(cm2min)

The pressure is increasing at a rate of approximately 0.457 kg/(cm2min).

Answer:

[tex] 0.609 \leq \sigma \leq 0.772[/tex]  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Step-by-step explanation:

1) Data given and notation  

s=0.68 represent the sample standard deviation  

[tex]\bar x =98.90[/tex] represent the sample mean  

n=98 the sample size  

Confidence=90% or 0.90  

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 mean or variance 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.  

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .  

2) Calculating the confidence interval  

The confidence interval for the population variance is given by the following formula:  

[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]  

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:  

[tex]df=n-1=98-1=97[/tex]  

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:  

[tex]\chi^2_{\alpha/2}=120.990[/tex]  

[tex]\chi^2_{1- \alpha/2}=75.282[/tex]  

And replacing into the formula for the interval we got:  

[tex]\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}[/tex]  

[tex] 0.371 \leq \sigma^2 \leq 0.596[/tex]  

Now we just take square root on both sides of the interval and we got:  

[tex] 0.609 \leq \sigma \leq 0.772[/tex]  

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Answer:

C. .556

Step-by-step explanation:

Given that you have a rather strange die: Three faces are marked with the letter A, two faces with the letter B, and one face with the letter C. You roll the die until you get a B.

Probability of getting A = [tex]\frac{3}{6}[/tex]

Probability of getting B = [tex]\frac{2}{6}[/tex]

Probability of getting c = [tex]\frac{1}{6}[/tex]

Each throw is independent of the other

the probability that the first B appears on the first or the second roll

= P(B) in I throw +P(B) in II throw

= P(B) in I throw + P(either A or C) in I throw*P(B) in II throw

=[tex]\frac{2}{6}+\frac{4}{6}*\frac{2}{6}\\=\frac{5}{9} \\=0.556[/tex]

Option c is right