What is the multiplicative rate of change of the function? One-fifth Two-fifths 2 5

Ephemeral services corporation (esco) knows that nine other companies besides esco are bidding for a $900,000 government contrac

EastWind [94]

Answer:

c. $100,000

Step-by-step explanation:

Calculation of the expected net profit of Ephemeral services corporation

Since we are been told that 9 other companies besides esco are as well bidding for the $900,000 government contract, it means we have to find the expected net profit by dividing 1 by 9×$900,000 .Thus ESCO can only expect to cover its sunk cost.

Hence ,

E(X) = (1/9) × $900,000

E(X)=0.111111111×$900,000

E(X)= $100,000

Therefore the expected net profit would be $100,000

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1 year ago

julie jot 21 of the 23 questions of her math test correct she got 29 of the 32 questions on her science test correct on which te

Alenkasestr [34]

$Make\ a\ common\ denominator:\\\\\frac{21}{23}\ \ \ \ \ \ \ \ \frac{29}{32}\\\\\frac{21\cdot32}{23\cdot32}\ \ \ \ \ \ \ \frac{29\cdot23}{32\cdot23}\\\\\frac{672}{736}\ \ \ > \ \ \ \frac{667}{736}$

5 0

1 year ago

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To plan the budget for next year a college must update its estimate of the proportion of next year's freshmen class that will ne

Naily [24]

Answer:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ 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 applicants who request financial aid is significantly higher than 0.35

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=67 represent the applicants who request financial aid

$\hat p=\frac{67}{150}=0.447$ estimated proportion of applicants who request financial aid

$p_o=0.35$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of applicants who request financial aid is higher than 0.35.:

Null hypothesis: $p\leq 0.35$

Alternative hypothesis: $p > 0.35$

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

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

Calculate the statistic

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

$z=\frac{0.447 -0.35}{\sqrt{\frac{0.35(1-0.35)}{150}}}=2.491$

Statistical decision

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

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

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

$p_v =P(z>2.491)=0.0064$

So the p value obtained was a very low value and using the significance level assumed $\alpha=0.05$ we have $p_v$ 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 applicants who request financial aid is significantly higher than 0.35

5 0

1 year ago

Presley is marking the lines for a soccer field. She needs to form a right triangle near the goal. She measures the lengths of h

Hunter-Best [27]

Answer:it’s b

Step-by-step explanation:

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1 year ago

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Coupons driving visits. A store randomly samples 603 shoppers over the course of a year and nds that 142 of them made their visi

s2008m [1.1K]

Answer:

The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)

Step-by-step explanation:

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

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 603, \pi = \frac{142}{603} = 0.2355$

95% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694$

The 95% confidence interval for the proportion of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694)

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1 year ago