A tire manufacturer believes that the life of its tires follow a normal distribution with a mean of 46,000 miles and a standard deviation of 4,000 miles. What mileage can he guarantee each tire to last so that 99% of the tires last longer than the guaranteed lifetime?

Guarantee of Tyre = 36,680 miles

Step-by-step explanation:

Given:

Mean = 46,000 miles

Standard deviation = 4,000 miles

Computation:

Assume;

x = guaranteed lifetime

P(X > x) = 99%

So,

P(X > x) = 0.99

So,

P(Z > z) = 0.99

So,

1 - P(Z > z )  = P(Z < z)

P(Z < z) = 0.01

So, in z table value of z = -2.33

x = μ + [z x σ]

x = 46,000 + [-2.33 x 4000]

x = 36,680

Guarantee of Tyre = 36,680 miles

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!