Laboratory tests show that the lives of light bulbs are normally distributed with a mean of 750 hours and a standard deviation of 75 hours. Find the probability that a randomly selected light bulb will last between 825 and 900 hours

P(825 to 900 hours) = 0.4772-0.3413=0.1359

Step-by-step explanation:

We are given that the life-span of the bulbs are normally distributed.

First we need to find the z-values, where

z(x)=(x-mean)/standard deviation

=(x-750)/75

therefore

z(825)=(825-750)/75=1

z(900)=(900-750)/75=2

Now we need to find the probabilities of the given z-values, from a probability table or from software.

P(z=1) = 0.3413

P(z=2)=0.4772

So probability that the life-span falls between 825 and 900 hours would be P(825 to 900 hour) = 0.4772-0.3413=0.1359

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!

answer: options

step-by-step explanation: