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2 years ago

14

Helppppppppppppppppppppppppp ASAP

1 answer:

S_A_V [24]2 years ago

3 0

Answer:

4.8

Step-by-step explanation:

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2 years ago

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Factor completely 10x^5+4x^4+8x^3

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2 years ago

Given the figure below, find the values of x and z. 80° 0 (15x + 40)

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2 years ago

The most hard question of all time that i need help :(

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2 years ago

The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches. Men the same

masha68 [24]

Answer: The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.

Step-by-step explanation:

Let x and y area the random variable that represents the heights of women and men.

Given : The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard deviation 2.7 inches.

i.e. $\mu_1 = 64$ $\sigma_1=2.7$

Since , $z=\dfrac{x-\mu}{\sigma}$

Then, z-score corresponds to a woman 6 feet tall (i.e. x=72 inches).

[∵ 1 foot = 12 inches , 6 feet = 6(12)=72 inches]

$z=\dfrac{72-64}{2.7}=2.96296296\approx2.96$

Men the same age have mean height 69.3 inches with standard deviation 2.8 inches.

i.e. $\mu_2 = 69.3$ $\sigma_2=2.8$

Then, z-score corresponds to a man 5'10" tall (i.e. y =70 inches).

[∵ 1 foot = 12 inches , 5 feet 10 inches= 5(12)+10=70 inches]

$z=\dfrac{70-69.3}{2.8}=0.25$

∴ The z-scores for a woman 6 feet tall is 2.96 and the z-scores for a a man 5'10" tall is 0.25.

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2 years ago