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DiKsa [7]
3 years ago
14

The owner of an electronics store received a shipment of MP3 players at a cost of $64.25 each. If he sells the MP3 players for $

99.99 each, what is the percent of markup? Round to the nearest whole percent.
A. 6%
B. 36%
C. 56%
D. 62%
Mathematics
1 answer:
Lostsunrise [7]3 years ago
4 0
99.99 - 64.25= 35.74
35.74/64.25= 0.556=55.6% or 56% rounded
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Step-by-step explanation:

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2 years ago
Autry made 45% of the shots he took during the basketball game. If he attempts 20 shots how many times did he score?
ratelena [41]

Answer:

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3 years ago
Suppose a sample of 972972 tenth graders is drawn. Of the students sampled, 700700 read above the eighth grade level. Using the
max2010maxim [7]

Answer:

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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

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For this problem, we have that:

972 students, 700 read above the eight grade level. We want the confidence interver for the proportion of those who read at or below the 8th grade level. 972 - 700 = 272, so n = 972, \pi = \frac{272}{972} = 0.28

85% confidence level

So \alpha = 0.15, z is the value of Z that has a pvalue of 1 - \frac{0.15}{2} = 0.925, so Z = 1.44.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 - 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.259

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.28 + 1.44\sqrt{\frac{0.28*0.72}{972}} = 0.301

The 85% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.259, 0.301).

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3 years ago
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kiruha [24]

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3 years ago
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