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2 years ago

8

What was the price elasticity of demand for iPhones in 2007 prior to the price reduction if the percentage change in price was 4

0%?

1 answer:

Greeley [361]2 years ago

8 0

For each dollar take 40 cents of and u will get the answer

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A box contains two defective Christmas tree lights that have been inadvertently mixed with eight nondefective lights. If the lig

REY [17]

Possibility 1: nondefective-defective-defective
possibility 2: defective-nondefective-defective

possibility 1: 8/10x 2/9 x 1/8 = 16/720=1/45
possibility 2: 2/10 x 8/9 x 1/8= 16/720=1/45

answer: 2/45

6 0

2 years ago

A sample of 900 college freshmen were randomly selected for a national survey. Among the survey participants, 372 students were

Rufina [12.5K]

Answer:

a) $ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

b) $0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

Step-by-step explanation:

1) Data given and notation

n=900 represent the random sample taken

X=372 represent the students were pursuing liberal arts degrees

$\hat p=\frac{372}{900}=0.413$ estimated proportion of students were pursuing liberal arts degrees

$\alpha=0.01$ represent the significance level

z would represent the statistic (variable of interest)

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

p= population proportion of students were pursuing liberal arts degrees

2) Solution to the problem

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

The margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

If we replace we have:

$ME=2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.0423$

And replacing into the confidence interval formula we got:

$0.413 - 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.371$

$0.413 + 2.58 \sqrt{\frac{0.413(1-0.413)}{900}}=0.455$

And the 99% confidence interval would be given (0.371;0.455).

We are confident (99%) that about 37.1% to 45.5% of students were pursuing liberal arts degrees.

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2 years ago

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First subtract 1.75n-2n-1.25n then you subtract 75 80 and 90 then you divide by n

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2 years ago

A sportswriter makes his pre-season picks for the top ten teams finish. If there are forty teams, how many different lists could

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$\bf _nC_r=\cfrac{n!}{r!(n-r)!}\qquad 
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1 year ago

Write a number that is greater than 4.508 but less than 4.512

Lelechka [254]

4.509 is greater that 4.508 and less than 4.512

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2 years ago

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