Ali and Bruno both run the 400 meter dash for their track team. Their coach frequently records their finishing times (in seconds), which are summarized in the table below. Mean Standard deviation
Ali \mu_A=59μ
A

=59mu, start subscript, A, end subscript, equals, 59 \sigma_A=4.8σ
A

=4.8sigma, start subscript, A, end subscript, equals, 4, point, 8
Bruno \mu_B=50μ
B

=50mu, start subscript, B, end subscript, equals, 50 \sigma_B=3.6σ
B

=3.6sigma, start subscript, B, end subscript, equals, 3, point, 6
Both distributions are approximately normal. Suppose we choose a random 400 meter race and calculate the difference between their times. It's reasonable to assume that their times are independent.
Find the probability that the Ali's finishing time is faster than Bruno's.
You may round your answer to two decimal places.
P\left(\text{Ali faster}\right)\approx

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!

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation: