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In-s [12.5K]
3 years ago
7

Which rate is the lowest price?

Mathematics
2 answers:
m_a_m_a [10]3 years ago
6 0

a. 1.15

b. 6.20 divided by 4 = 1.55

c. 5 divided by 4 = 1.25

d. 5.50 divided by 5 = 1.1


D is correct

Alexandra [31]3 years ago
6 0

A)1.15 each

B)1.55

C)1.25

D)1.10

The answer is D because 5.50 divided by 5= 1.10.

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Step-by-step explanation:

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Round to the nearest hundredth
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Hey there,
1st hat = x
2nd hat = 2x
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The American Management Association is studying the income of store managers in the retail industry. A random sample of 49 manag
VashaNatasha [74]

Answer:

a) The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

b) The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

Step-by-step explanation:

Question a:

We have to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a p-value of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

M = 1.96\frac{2050}{\sqrt{49}} = 574

The lower end of the interval is the sample mean subtracted by M. So it is 45420 - 574 = $44,846.

The upper end of the interval is the sample mean added to M. So it is 45420 + 574 = $45,994.

The 95% confidence interval for the income of store managers in the retail industry is ($44,846, $45,994), having a margin of error of $574.

Question b:

The interval mean that we are 95% sure that the true mean income of all store managers in the retail industry is between $44,846 and $45,994.

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Step-by-step explanation:

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I've done this question: Its batch 5

Step-by-step explanation:

It has to be all equal proportions

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