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Leya [2.2K]
3 years ago
7

Match the following equation to the correct situation.

Mathematics
1 answer:
uranmaximum [27]3 years ago
8 0

Answer:

The equation given in the question matches the situation C.

Step-by-step explanation:

A. The amount that Cindy paid for books was 15% of p i.e. it is only \frac{15p}{100} = 0.15p

B. The amount is given to a charity after saving 15% of p. That means if p is the amount of charity, then there were a savings of 15% of p.

Therefore, the total charity payment after savings = p - 15% of p = p - \frac{15p}{100} = p - 0.15p = 0.85p

C. The amount that Sandy paid for a shirt with 15% tax on p.

Therefore, if the price of a shirt is p, then total payment of Sandy after tax is = p + 15% of p = (p + \frac{15p}{100}) = p + 0.15p = 1.15p

Hence, the equation given in the question matches the situation C. (Answer)

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C. Point is a defined term

Step-by-step explanation:

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2 7/12 hours

X=101/2 - ( 3+4+1/4+2/3)

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3 years ago
At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro
Gnom [1K]

Answer:

A sample of 1068 is needed.

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}}

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.

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of \pi, so we use the worst case scenario, which is \pi = 0.5

Then

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

0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.03\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.03}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}

n = 1067.11

Rounding up

A sample of 1068 is needed.

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