GUYS PLS HELP THIS IS MY LAST QUESTION

Mathematics

1 answer:

erastovalidia [21]3 years ago

3 0

Answer:

Let Q be (x,y). Then AQ = 4QB; or (x - 2) = 4(4 - x) and (y + 1) = 4(2 - y);

x - 2 = 16 - 4x; 5x = 18 [ x = 3.6. y + 1 = 8 - 4y; 5y = 7, y = 1.4, So, Q(3.6, 1.4)

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The heights of American men have an approximately bell-shaped distribution with a mean of 70 inches (5' 10") and standard deviat

Katyanochek1 [597]

Answer:

0.9973 is the percentage of American men whose height is between 61 and 79 inches.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 70 inches

Standard Deviation, σ = 3 inches

We are given that the distribution of heights of American men is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(height is between 61 and 79 inches)

$P(61 \leq x \leq 79) = P(\displaystyle\frac{61 - 70}{3} \leq z \leq \displaystyle\frac{79-70}{3}) = P(-3 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -3)\\= 0.9987 - 0.0014 = 0.9973= 99.73\%$