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Airida [17]
2 years ago
6

Work out the area of the circle and leave it in cm plz

Mathematics
2 answers:
Marat540 [252]2 years ago
5 0

Answer:

132.7495 cm^2

Step-by-step explanation:

area of circle=πr^2

radius=6.5

6.5^2

42.25

42.25*3.142

132.7495

miskamm [114]2 years ago
3 0

Answer:

132.7495 cm

Step-by-step explanation:

13/2=6.5

6.5*6.5=42.25

42.25*3.142=132.1495

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Answer: \dfrac{K}{2} .

Step-by-step explanation:

As we know ,

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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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