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True [87]
3 years ago
8

Describe the relationship between n and 4 that will make the value of the expression 7x n/4 greater than 7

Mathematics
1 answer:
vladimir1956 [14]3 years ago
6 0
The relationship between n and 4 is n>4

We have,
     
    7\frac{n}{4} > 7

The number 7 which is in multiplication on left hand side of the expression, will come in denominator on right hand side of the expression.

⇒ \frac{n}{4} > \frac{7}{7}

⇒ \frac{n}{4} > 1
Again 4 in denominator from left hand side will come in numerator on right hand side
⇒ n > 4
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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
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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.

5 0
3 years ago
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