1500 people subscribed Sunday edition and 250 people subscribed only sunday edition.
<u>Solution:</u>
Given, A survey of 2500 subscribers to a certain news paper revealed that 2250 people subscribe to the daily morning edition
And 1250 subscribe to both the daily and the sunday editions.
We have to find how many subscribe to the Sunday edition? how many subscribe to the Sunday edition only?
Let x denote the number who subscribe to the Sunday edition.
Then the addition rule with overlap tells us that
Those who subscribe daily edition + those who subscribe sunday edition - those who subscribe both daily and sunday edition = total subscribers in survey
2250 + x – 1250 = 2500
1000 + x = 2500
x = 1500
so, 1500 subscribe to the Sunday edition and 1500 – 1250 = 250 subscribe to the Sunday edition only.
Hence, 1500 people subscribed Sunday edition and 250 people subscribed only sunday edition.
The product of -8 x 9 x 2= -144
Answer:
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% !
The best answer to the question that is being stated above would be that A and B are both dependent. If I was told that P(A | B) = P (B | A), then the true statement is that both variables inside would be dependent upon each other, no matter if you exchange their places with one another.
C. f(x)= log 1/x^2 , c=/ 0
