The number of three-digit numbers with distinct digits that 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 .

a)120

b)6.67%

Step-by-step explanation:

Given:

No. of digits given= 6

Digits given= 1,2,3,5,8,9

Number to be formed should be 3-digits, as we have to choose 3 digits from given 6-digits so the no. of combinations will be

6P3= 6!/3!

      = 6*5*4*3*2*1/3*2*1

      =6*5*4

      =120

Now finding the probability that both the first digit and the last digit of the three-digit number are even numbers:

As the first and last digits can only be even

then the form of number can be

a)2n8 or

b)8n2

where n can be 1,3,5 or 9

4*2=8

so there can be 8 three-digit numbers with both the first digit and the last digit even numbers

And probability = 8/120

                          = 0.0667

                          =6.67%

The probability that both the first digit and the last digit of the three-digit number are even numbers is 6.67% !

1.

2.

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

range =5

step-by-step explanation:

to find the range, we take the largest value and subtract the smallest value.

the largest value is 5

the smallest value is 0

the range is 5-0   =5