The time for an oil change is normally distributed with mean μ = 11.5 minutes and standard deviation LaTeX: \sigma σ= 1.2 minutes. For what percentage of cars will the oil change take between 9 and 13 minutes?

87.572%

Step-by-step explanation:

We start by calculating the z-scores for each of the minutes

Z-score = (x-mean)/SD

Where x is the raw score, mean = 11.5 and standard deviation is 1.2

For 9 minutes

z-score = (9-11.5)/1.2 = -2.08

For 13 minutes

z-score = (13-11.5)/1.2 = 1.25

So the probability we want to calculate is;

P( -2.083 < x < 1.25)

we can use the standard normal distribution table to get this

The probability will be;

P(x < 1.25) - P (-2.083< x) = 0.87572

This represents a percentage of 87.572%

the answer is d

step-by-step explanation:

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl