Given a normal distribution with a mean of 120 and a standard deviation of 12, what percentage of values is within the interval 96 to 144

0.9545

Step-by-step explanation:

Given that :

Mean (m) = 120

Standard deviation (s) = 12

what percentage of values is within the interval 96 to 144

P(96 < x < 144)

P(x < 96)

obtain the standardized score (Z) :

Z = (x - m) / s

Z = (96 - 120) / 12 = - 2

p(Z < - 2) = 0.02275 ( Z probability calculator)

P(x < 144)

obtain the standardized score (Z) :

Z = (x - m) / s

Z = (144 - 120) / 12 = 2

p(Z < - 2) = 0.97725 ( Z probability calculator)

p(Z < - 2) - p(Z < 2) = 0.97725 - 0.02275 = 0.9545

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!

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.