Which of the following random variables is not discrete?

a) the number of classes taken in one semester by a student

b) the annual rainfall in a city

c) the attendance at a football game

d) the number of patients treated at an emergency room in a day

b) The annual rainfall in a city

Step-by-step explanation:

Remember, a discrete variable is one that can only take a finite number of values between any two values of a characteristic and a continuous variable is one that can take an infinite number of values between any two values of a characteristic.

a) Observe that the variable x='classes taken in one semester' can take the values 0,1,2,...,n.

Then the variable x is discrete

b) Observe that the variable x='annual rainfall in a city' can take the values 2in, 1.6in, 5.1 in, 0.1in

Then, the variable x can be take a infinite number of values between two number. So x isn't a discrete variable.

c) The variable x='attendance at a football game' can take the values 3000,5000... n. And never will be a decimal number because There cannot be a personal decimal number. Therefore, x is a discrete variable.

d) The variable x='patients treated at an emergency room in a day' can take the values 1,2,3,...,n. And never will be a decimal number because There cannot be a personal decimal number. Therefore, x is a discrete variable.

step-by-step explanation:

x^2 -12x + 36 + y^2 = 36

x^2 + y^2 = 12x

r^2 = 12rcos(theta)

r = 12 cos(theta)

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!