Taking the square root of both sides gives two possible cases,
or
Recall that
If and
, we have
so in the equations above, we can write
Then in the first case,
(where is any integer)
and in the second,
Then the solutions that fall in the interval are
Answer:
(a)There are 500 integers from 1 through 1,000 are multiples of 2.
There are 111 integers from 1 through 1,000 are multiples of 9.
(b)0.611
Step-by-step explanation:
Given the set of Integers from 1 through 1000
The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.
To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,
where first term, a =2, common difference, d =2
Nth term of an A.P,
Similarly,
Given the set of Integers from 1 through 1000
The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.
To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,
where first term, a =9, common difference, d =9
Nth term of an A.P,
(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.
There are a total of 1000 numbers between from 1 through 1,000
n(S)=1000
n(Multiples of 2)=500
n(Multiples of 9)=111
Therefore:
