11.1.30
How many three-digit even numbers are possible if the leftmost digit cannot be zero?

There are 450 three-digit even numbers where the leftmost digit cannot be zero

The smallest even three-digit is 100, and the largest even three-digit is 998.

The difference between consecutive even numbers is 2.

So, the count of even numbers is calculated using the following arithmetic progression formula

Where:

-- the last term

-- the first term

-- the common difference

So, we have:

Subtract 100 from both sides

Divide both sides by 2

Add 1 to both sides

Hence, there are 450 three-digit even numbers

Read more about arithmetic progressions at:

link

The total number of 3 digit even numbers is 450

Step-by-step explanation:

Step 1 :

We need to determine the  number of 3 digit numbers which is even and the leftmost digit is not zero .

The  3 digits with left most digit not equal to zero starts from 100 and goes up to 998

If we consider only the even 3 digit numbers in this interval, this would form an arithmetic progression with the first number a = 100 and the common difference d = 2.

Step 2:

The last number l  in this series is 998 .

So we have

a = 100

l = 998

d = 2

The nth term in the arithmetic progression is given by a + (n-1) d

so substituting the above values we get

100 + (n-1) 2 = 998

2n = 900  => n = 450

Step 3 :

Answer :

The total number of 3 digit even numbers is 450

question

what is the value of the third quartile of the data set represented by this box plot? a. 19 b. 21 c. 26 d. 29

answer

c

step-by-step explanation: