<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7B-140%7D%7B%5Csqrt%7B28%7D%20%7D" id="TexFormula1" title="\sqrt{\frac{-140}

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Alex777 [14]

2 years ago

{\sqrt{28} }" alt="\sqrt{\frac{-140}{\sqrt{28} }" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Zanzabum2 years ago

8 0

This is the answer

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Write each percent as a fraction<br> 10. 56<br> 564%

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10.56 as a fraction is 264 / 25
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krek1111 [17]

Answer:

So we use the t-distribution to compute a confidence interval for the average age of people using online dating services

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Confidence interval for a mean.

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It takes 10 minutes to fill a 60-gallon bathing pool. What is the average rate in gallons per hour

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Trava [24]

Answer:

(a) $\log_3(\dfrac{81}{3})=3$

(b) $\log_5(\dfrac{625}{25})=2$

(d) $\log_4(\dfrac{64}{16})=1$

(e) $\log_6(36^4)=8$

(f) $\log(100^3)=6$

Step-by-step explanation:

Let as consider the given equations are $\log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?$ .

(a)

$\log_3(\dfrac{81}{3})=\log_3(27)$

$\log_3(\dfrac{81}{3})=\log_3(3^3)$

$\log_3(\dfrac{81}{3})=3$ $[\because \log_aa^x=x]$

(b)

$\log_5(\dfrac{625}{25})=\log_5(25)$

$\log_5(\dfrac{625}{25})=\log_5(5^2)$

$\log_5(\dfrac{625}{25})=2$ $[\because \log_aa^x=x]$

(c)

$\log_2(\dfrac{64}{8})=\log_2(8)$

$\log_2(\dfrac{64}{8})=\log_2(2^3)$

$\log_2(\dfrac{64}{8})=3$ $[\because \log_aa^x=x]$

(d)

$\log_4(\dfrac{64}{16})=\log_4(4)$

$\log_4(\dfrac{64}{16})=1$ $[\because \log_aa^x=x]$

(e)

$\log_6(36^4)=\log_6((6^2)^4)$

$\log_6(36^4)=\log_6(6^8)$

$\log_6(36^4)=8$ $[\because \log_aa^x=x]$

(f)

$\log(100^3)=\log((10^2)^3)$

$\log(100^3)=\log(10^6)$

$\log(100^3)=6$ $[\because \log10^x=x]$

5 0

3 years ago