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Dominik [7]
3 years ago
8

Mica bought a new smart phone for $258.13. That was the price after his 18.25% discount. What was the original price of the phon

e?
Mathematics
1 answer:
julia-pushkina [17]3 years ago
4 0

The original price of phone is $ 315.76

<em><u>Solution:</u></em>

Given that Mica bought a new smart phone for $ 258.13.

That was the price after his 18.25% discount

<em><u>To find: original price of phone</u></em>

From given question,

price after discount = $ 258.13

discount = 18.25 %

Let "x" be the original price of phone

price after discount = original price - discount

258.13 = x - 18.25 \% \text{ of } x

258.13 = x - \frac{18.25}{100}x

258.13 = x(1 - \frac{18.25}{100})

258.13 = x(1 - 0.1825)\\\\258.13 = x( 0.8175)\\\\x = \frac{258.13}{0.8175}\\\\x = 315.76

Thus original price of phone is $ 315.76

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Answer:

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

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}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

For this problem, we have that:

We dont know the true proportion, so we use \pi = 0.5, which is when we are are going to need the largest sample size.

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.

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.04\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.04}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2

n = 600.25

Rounding up

The minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.02\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.02}

(\sqrt{n})^2 = (\frac{1.96*0.5}{0.02})^2

n = 2401

The minimum sample size is 2401.

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