Explain how the Quotient of Powers was used to simplify this expression.

2 answers:

5 0

The correct answer on how the Quotient of Powers was used to simplify the expression is D. By finding the quotient of the bases to be one-fifth and simplifying the expression. Thank you for posting your question. I hope this answer helped you. Let me know if you need more help.

Ymorist [56]3 years ago

5 0

5^4
------ = 52
25

5^4
------ = 52 oh i think really this 52 mean to be 5^2
5^2

5^(4-2) = 5^2

5^2 = 5^2

hope helped

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Goal is to make $100. What % of the goal was met if raised $350?

slega [8]

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<h3>A = m ÷ t × 100</h3><h3>= $100 ÷ $350 × 100</h3><h3>= <u>28.6%</u></h3><h3>THE GOAL WAS SURPASSED BY 28.6%</h3>

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7 0

3 years ago

Need help pls<br>thank you in advance

ira [324]

Answer:

<u />

$\textsf{Mean}\:\overline{X}=\sf \dfrac{\textsf{sum of all the data values}}{\textsf{total number of data values}}$

$\implies \sf Mean\:(Nilo)=\dfrac{5+6+14+15}{4}=\dfrac{40}{4}=10$

$\implies \sf Mean\:(Lisa)=\dfrac{8+9+11+12}{4}=\dfrac{40}{4}=10$

<h3><u>Standard Deviation</u></h3>

$\displaystyle \textsf{Standard Deviation }s=\sqrt{\dfrac{\sum X^2-\dfrac{(\sum X)^2}{n}}{n-1}}$

$\begin{aligned}\displaystyle \textsf{Standard Deviation (Nilo)} & =\sqrt{\dfrac{(5^2+6^2+14^2+15^2)-\dfrac{(5+6+14+15)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{482-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{82}{3}}\\\\& = 5.23\end{aligned}$

$\begin{aligned}\displaystyle \textsf{Standard Deviation (Lisa)} & =\sqrt{\dfrac{(8^2+9^2+11^2+12^2)-\dfrac{(8+9+11+12)^2}{4}}{4-1}}\\\\& = \sqrt{\dfrac{410-\dfrac{40^2}{4}}{3}}\\\\& = \sqrt{\dfrac{10}{3}}\\\\& = 1.83\end{aligned}$

<h3><u>Summary</u></h3>

Nilo has a mean score of 10 and a standard deviation of 5.23.

Lisa has a mean score of 10 and a standard deviation of 1.83.

The <u>mean</u> scores are the <u>same</u>.

Nilo's standard deviation is higher than Lisa's. Therefore, Nilo's test scores are more <u>spread out</u> that Lisa's, which means Lisa's test scores are more <u>consistent</u>.

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1 year ago

State the domain and the range of each relation. Then determine whether the relation is a function.

IrinaVladis [17]

Answer:

$Domain:$ { $-6,3,4$ }