The probability
The number of three-digit numbers with distinct digits that can be formed using the digits 1, 2, 3,5,8, and 9 is _. The probability that both the first digit and the last digit of the three-digit number are even numbers is _.

Probability of distinct 3-digits is 1/6.

Probability of both the first digit and the last digit are even numbers is 1/5.

Step-by-step explanation:

Given set of numbers {1,2,3,5,8,9} = 6

Total number of 3-digits can be formed = 654 = 120.

Probability  of the three-digit numbers with distinct digits :

The number of 3-digit numbers with distinct digits = (654) / (321) = 20.

P(distinct 3-digits) =  no. of distinct 3-digits / Total no.of 3-digits

⇒ 20/120

⇒ 1/6

Probability that both the first digit and the last digit of the three-digit number are even numbers :

The even numbers in the set are 2 and 8 = 2 possibilities.

The number of 3-digits with 1st and 3rd digit are even = (262) = 24.

P(1st and 3rd digits even)= no. of 1st and 3rd digit even/Total no. of 3-digits.

⇒ 24/120

⇒ 1/5

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!

b. ol

step-by-step explanation:

i got the answer right on edg.

congruent: (geometry) (of figures) identical in form; coinciding exactly when superimposed.