If adult females are randomly selected, find the probability that they have pulse rates with a mean between 70beats per minute and 82beats per minute.

This question is incomplete.

Complete Question

Assume that females have pulse rates that are normally distributed with a mean of u = 76.0 beats per minute and a standard deviation of 12.5 beats per minute. If 1 adult female is randomly selected, find the probability that her pulse rate is between 70 beats per minute and 82 beats per minute. The probability is (Round to four decimal places as needed.)

0.3688

Step-by-step explanation:

The formula for calculating a z-score when given a random sample of numbers is is z = (x-μ)/σ/√n

where x is the raw score,

μ is the population mean, and

σ is the population standard deviation.

mean = 76.0 beats per minute and a standard deviation = 12.5 beats per minute.

n = number of random samples = 1

For 70 beats per minute

z = 70 - 76/12.5/√1

z = -0.48

Probability value from Z-Table:

P(x = 70) = 0.31561

For 82 beats per minute.

z = 82 - 76/12.5/√1

z = 0.48

Probability value from Z-Table:

P(x = 82) = 0.68439

The probability that her pulse rate is between 70 beats per minute and 82 beats per minute.

P(x = 82) - P(x = 70)

0.68439 - 0.31561

= 0.36878

≈ 0.3688

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3