Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than 1.99 and draw a sketch of the region.

Step-by-step explanation:

To find this probability, we shall be using the z-score route

Mathematically ;

z-score = (x -mean)/SD

From the question, x = 1.99, mean = 0 and SD = 1

So z = (1.99-0)/1 = 1.99

So the probability we want to calculate is;

P(z<1.99)

This value can be obtained from the standard normal distribution table.

P(z < 1.99) = 0.9767

The sketch of the region is as shown as in the attachment.

if a = 1,   and b = 2, then the answer is 47.99999999. if you round that to the nearest whole number, the answer is 48.

option 3 . (5x + 1) - (-10x + 6)

step-by-step explanation:

option 1 :

(4x + 3) + (-9x + 4)

or 4x+3 -9x + 4

or -5x + 7

option 2

(x - 2) - (6x - 9)

or x-2-6x + 9

or -5x + 7

option 3

(5x+ + 6)

or 5x + 1 + 10x -6

or 15x -5

option 4

-4x+2x+7-3x

-7x + 2x + 7

-5x + 7

hence only option 3 did   not satisfy the given simplify expression.