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Taya2010 [7]
3 years ago
13

The price of one share of Starbucks declined five dollars per day for four days in a row. How much did the price of one share ch

ange in total after the four days?
Mathematics
1 answer:
myrzilka [38]3 years ago
5 0

Answer:

Each share lost $20 in value over 4 days time

Step-by-step explanation:

$5 loss for 4 days

5 x 4 = 20

$20 loss in 4 days.

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Answer:

2

Step-by-step explanation:

well less than 3

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Find the new price after the markdown given. Round to the nearest cent if necessary. $5,250 marked down 15%
PIT_PIT [208]

Marked down 15% means the new price would be 85% of the original price ( 100%-15% - 85%)

Multiply the original price by 85%:

5250 x 0.85 = 4462.50

The new price is $4,462.50

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Step-by-step explanation:

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Can someone please help me with the assignment, I'm almost finish!
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Answer:

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In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent (Pew Research, June 5, 2014). An independe
enot [183]

Answer:

a)

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2, in which z is related to the confidence level.

b) A sample size of 991 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}}

In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent

This means that \pi = 0.16

a. What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03 (round up to the next whole number).

This is n for which M = 0.03. So

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

0.03 = z\sqrt{\frac{0.16*0.84}{n}}

0.03\sqrt{n} = z\sqrt{0.16*0.84}

\sqrt{n} = \frac{z\sqrt{0.16*0.84}}{0.03}

(\sqrt{n})^2 = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2, in which z is related to the confidence level.

Question b:

99% confidence level,

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

n = (\frac{2.575\sqrt{0.16*0.84}}{0.03})^2

n = 990.2

Rounding up

A sample size of 991 is needed.

8 0
3 years ago
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