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IRISSAK [1]
3 years ago
8

5^(2x-1)+5^(x+1)=250 how do you solve? thank you

Mathematics
1 answer:
zepelin [54]3 years ago
8 0

Answer:

<em>x = 2</em>

Step-by-step explanation:

<u>Exponential Equations</u>

Solve:

5^{2x-1}+5^{x+1}=250

Separate each exponential:

5^{2x}5^{-1}+5^{x}5^{1}=250

Operating:

\displaystyle \frac{5^{2x}}{5}+5^{x}5=250

Multiplying by 5:

5^{2x}+25\cdot5^x=1250

Rearranging:

5^{2x}+25\cdot5^x-1250=0

Recall that:

5^{2x}=(5^{x})^2

(5^{x})^2+25\cdot5^x-1250=0

Calling

y=5^{x}:

y^2+25y-1250=0

Factoring:

(y-25)(y+50)=0

There are two possible solutions:

y=25

y=-50

Since

y=5^{x}

y cannot be negative, thus:

5^{x}=25=5^2

The solution is:

x = 2

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Alecsey [184]

Answer:

CV=0.2 ---- dataset 1

CV = 7.2 --- dataset 2

Step-by-step explanation:

Given

A: 30500, 27500, 31200, 24000, 27100,28600, 39100, 36900, 35000, 21400, 37900, 27900, 18700,33100

B: 4.29, 4.88, 4.34, 4.17, 4.52, 4.80, 3.28, 3.79, 4.84, 4.77, 3.11

Required

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Calculate the mean

\mu = \frac{\sum x}{n}

\mu = \frac{30500+ 27500+31200+24000+ 27100+28600+ 39100+ 36900+ 35000+ 21400+ 37900+ 27900+ 18700+33100}{14}\mu = \frac{418900}{14}

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Next, calculate the standard deviation using:

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\sigma= \sqrt{\frac{(30500 - 29921.43)^2 +.................+ (18700- 29921.43)^2 + (33100- 29921.43)^2}{13}}

\sigma= \sqrt{\frac{487723571.42857}{14}}

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CV=\frac{5902.32}{29921.43}

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Calculate the mean

\mu = \frac{\sum x}{n}

\mu = \frac{4.29+ 4.88+ 4.34+ 4.17+ 4.52+ 4.80+ 3.28+ 3.79+ 4.84+ 4.77+ 3.11}{11}

\mu = \frac{46.79}{11}

\mu = 4.25

Next, calculate the standard deviation using:

\sigma = \sqrt{\frac{\sum(x - \mu)^2}{n}}

\sigma = \sqrt{\frac{(4.29 - 4.25)^2 + (4.88- 4.25)^2 +.........+ (3.11- 4.25)^2}{11}}

\sigma = \sqrt{\frac{3.859}{11}}

\sigma = \sqrt{0.35081818181}

\sigma = 0.593

So, the coefficient of variation is:

CV=\frac{\sigma}{\mu}

CV = \frac{4.25}{0.5903}

CV = 7.2 -- approximated

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