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Slav-nsk [51]
3 years ago
8

Find the value of 'a' if 2^2a×2^2a=128

Mathematics
1 answer:
masya89 [10]3 years ago
4 0
128=2⁷
2^2a x 2^2a=2⁷
2^(2a+2a)=2⁷
2^4a=2⁷

Therefore:
4a=7
a=7/4
a=1.75

Answer: a=1.75

To check
2²⁽⁷/⁴⁾ * 2²⁽⁷/⁴⁾ =2⁷/² * 2⁷/²=2¹⁴/²=2⁷=128
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Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the pop
Andreyy89

Answer:

The smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.

Step-by-step explanation:

The complete question is:

The mean salary of people living in a certain city is $37,500 with a standard deviation of $2,103. A sample of n people will be selected at random from those living in the city. Find the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income. Round your answer up to the next largest whole number.

Solution:

The (1 - <em>α</em>)% confidence interval for population mean is:

CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

The margin of error for this interval is:

MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

The critical value of <em>z</em> for 90% confidence level is:

<em>z</em> = 1.645

Compute the required sample size as follows:

MOE=z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}

      n=[\frac{z_{\alpha/2}\cdot\sigma}{MOE}]^{2}\\\\=[\frac{1.645\times 2103}{500}]^{2}\\\\=47.8707620769\\\\\approx 48

Thus, the smallest sample size n that will guarantee at least a 90% chance of the sample mean income being within $500 of the population mean income is 48.

3 0
3 years ago
1/3×24 to the power of 2 times 30
svp [43]
1/3*(24^2)*30=5760

Any questions just ask. Thanks
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3 years ago
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