in the life of a car engine, calculatedin miles, is normally distributed, with a mean of 17,000 miels and a standard deviation of 16,500 miles, what should be the guarantee period if the company wants less than 2% of the engines to fail while under warranty g

the guarantee period should be less than 136010 miles

Step-by-step explanation:

From the given information;

Let consider Y to be the life of a car engine

with a mean μ = 170000

and a standard deviation σ = 16500

The objective is to determine what should be the guarantee period T if the company wants less than 2% of the engines to fail.

i.e

P(Y < T ) < 0.02

For the variable of z ; we have:

Now;

From Z table ;

At P(Z < -2.06) ≅ 0.0197  which is close to 0.02

Thus; the guarantee period should be less than 136010 miles

a 4/6 = 9/12   not proportional

step-by-step explanation:

to find which are not proportional, we can use cross products

if it is true, they are proportional.   if it is not true, they are not proportional.

a 4/6 = 9/12

4* 12 = 6*9

48 = 54

not this is not proportional  

b 2/3 = 12/18  

2* 18 = 3 * 12

36 = 36

this is proportional

c 5/4 = 10/8  

5*8 = 4*10

40 = 40

this is proportional

d 7/5 = 35/25

7*25 = 5 * 35

175 = 175

is proportional

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph