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kari74 [83]
3 years ago
10

Thesquare(5x² + 6xy)²is​

Mathematics
2 answers:
kari74 [83]3 years ago
8 0

Answer:

{25x}^{4}  + 60 {x}^{3} y + 36 {x}^{2}  {y}^{2}

Step-by-step explanation:

(5 {x}^{2} )^{2}  + 2 \times 5 {x}^{2}  \times 6xy + (6xy)^{2}

Gives the above answer

dybincka [34]3 years ago
4 0

Answer:

in the picture

Step-by-step explanation:

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Step-by-step explanation:

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The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.
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Answer:

(a) 0.2061

(b) 0.2514

(c) 0

Step-by-step explanation:

Let <em>X</em> denote the heights of women in the USA.

It is provided that <em>X</em> follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

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