Answer:
The three digit number = 951
Step-by-step explanation:
Let suppose the numbers are:
= abc
According to given condition:
a+ b + c = 15 -------------eq1
Also, given the difference between the first two digit = the difference between the last two digits:
==> l a-b l = l b-cl
==> (a-b) = (b-c)
==> (a+c) = 2b
Now we will substitue in eq1
==> a+ b + c = 15
==> 2b + b = 15
==> 3b = 15
Dividing both sides by 3 we get:
b =5
a + c = 2b
a+ c = 10
a = 10 -c ..........(2)
We know that"
(a-b) = (b-c)
==> a > b+c
==> a > 5 + c
==> 10 -c > 5 +c
==>5 > 2c
==> 2.5 > c
As c is an odd number so c will be equal to 1
c = 1
a = 10 -1
a = 9
The three digit number = 951
The hundred digit is greater than the sum of the tens and ones digits
i hope it will help you!
Answer:
The value of P(AUB) = 0.438
Step-by-step explanation:
Given:
P(A) = 0.36
P(B) = 0.2
P(A∩B) = 0.122
Find:
The value of P(AUB)
Computation:
P(AUB) = P(A) + P(B) - P(A∩B)
The value of P(AUB) = 0.36 + 0.2 - 0.122
The value of P(AUB) = 0.56 - 0.122
The value of P(AUB) = 0.438
To do these problems, you plug in the ‘n’ value given into the equation.
a. 6j - 3 : j = 4
6(4) - 3
24 - 3
21
b. 1/2b + 5 : b = 14
1/2(14) + 5
7 + 5
12
c. 8 + 4k : k = 3.5
8 + 4(3.5)
8 + 14
22
Answer:
3547.90
Step-by-step explanation:
