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Likurg_2 [28]
3 years ago
14

The distance between two points is 10 units, if the coordinates of one of the endpoints are (4, -7), find x if the coordinates o

f the other endpoint are (x, 1).​
Mathematics
1 answer:
PIT_PIT [208]3 years ago
8 0

Answer:

10

Step-by-step explanation:

let the distance = d

d² = (x2-x1)² + (y2-y1)²

=>

10²= (x-4)²+(1+7)²

100 = (x-4)²+64

(x-4)²=100-64

= 36

x-4 = √36

x-4=6

x= 6+4

x= 10

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3 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.

6 0
3 years ago
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