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Novay_Z [31]
3 years ago
6

Help! 10 Points not 5!

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
6 0

9514 1404 393

Answer:

  18

Step-by-step explanation:

Solve the first for a, then substitute into the second.

  a = 27/(5b^2)

Then ...

  (27/(5b^2))^2·b = 135

  27^2/(5^2·135) = b^3 = 27/125

  b = ∛(27/125) = 3/5

  a = 27/(5(3/5)^2) = 15

__

The expression of interest is ...

  a +5b = 15 + 5(3/5) = 15 +3

  a +5b = 18

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A population of values has a normal distribution with μ=204.9μ=204.9 and σ=81.9σ=81.9. You intend to draw a random sample of siz
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Answer:

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8. μ=75.9 and σ=0.7136

9. p=0.9452

Step-by-step explanation:

7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.

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\mu_x=\mu=204.9, \mu_x=sample \ mean

-The standard deviation,\sigma_x is calculated as:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968

8. For a random variable X.

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#The sample standard deviation is calculated as follows:

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9. Given the population mean, μ=135.7 and σ=88 and n=59

#We calculate the sample mean;

\mu_x=\mu=135.7

#Sample standard deviation:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566

#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:

P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452

Hence, the probability of a random sample's mean being greater than 117.4 is 0.9452

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