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Luda [366]
3 years ago
12

Consider the expression. (StartFraction 2 m Superscript negative 1 baseline n Superscript 5 Baseline Over 3 m Superscript 0 Base

line n Superscript 4 Baseline EndFraction) squared What is the value of the expression if m = –5 and n = 3? Negative StartFraction 24 Over 25 EndFraction Negative StartFraction 4 Over 25 EndFraction StartFraction 4 Over 25 EndFraction StartFraction 24 Over 25 EndFraction
Mathematics
2 answers:
suter [353]3 years ago
8 0

Answer:

The correct option is option (3) 4 ÷ 25.

Step-by-step explanation:

The expression in terms of <em>m</em> and <em>n</em> is:

F(m,n)=[\frac{2m^{-1}n^{5}}{3m^{0}n^{4}}]^{2}

Exponent rule of division:

a^{x}\div a^{y}=a^{x-y}

Compute the value of the expression for <em>m</em> = -5 and <em>n</em> = 3 as follows:

F(m,n)=[\frac{2m^{-1}n^{5}}{3m^{0}n^{4}}]^{2}

F(-5,3)=[\frac{2\csdot (-5)^{-1}\cdot (3)^{5}}{3\cdot (-5)^{0}\cdot (3)^{4}}]^{2}

             =\{\frac{2}{3}\times [(-5)^{-1-0}\times (3)^{5-4}}]\}^{2}\\\\=\{\frac{2}{3}\times \frac{-1}{5}\times 3\}^{2}\\\\=\{-\frac{2}{5}\}^{2}\\\\=\frac{4}{25}

Thus, the correct option is option (3) 4 ÷ 25.

Nata [24]3 years ago
6 0

Answer:

4/25

Step-by-step explanation:

I just did the test

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3 years ago
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OleMash [197]

Answer:

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Step-by-step explanation:

5 0
2 years ago
Suppose a 90% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$38,737, $50,46
JulsSmile [24]

Answer:

a. The point estimate was of $44,600.

b. The sample size was of 16.

Step-by-step explanation:

Confidence interval concepts:

A confidence interval has two bounds, a lower bound and an upper bound.

A confidence interval is symmetric, which means that the point estimate used is the mid point between these two bounds, that is, the mean of the two bounds.

The margin of error is the difference between the two bounds, divided by 2.

a. What is the point estimate of the mean salary for all college graduates in this town?

Mean of the bounds, so:

(38737+50463)/2 = 44600.

The point estimate was of $44,600.

b. Determine the sample size used for the analysis.

First we need to find the margin of error, so:

M = \frac{50463-38737}{2} = 5863

Relating the margin of error with the sample size:

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

\alpha = \frac{1 - 0.9}{2} = 0.05

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 pvalue of 1 - 0.05 = 0.95, so Z = 1.64.

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.

For this problem, we have that \sigma = 14300, M = 5863. So

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

5863 = 1.645\frac{14300}{\sqrt{n}}

5863\sqrt{n} = 1.645*14300

\sqrt{n} = \frac{1.645*14300}{5863}

(\sqrt{n})^2 = (\frac{1.645*14300}{5863})^2

n = 16

The sample size was of 16.

7 0
3 years ago
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Oduvanchick [21]
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2 years ago
Read 2 more answers
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Zinaida [17]

Answer:

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Step-by-step explanation:

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2 years ago
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