Given:
Three numbers in an AP, all positive.
Sum is 21.
Sum of squares is 155.
Common difference is positive.
We do not know what x and y stand for. Will just solve for the three numbers in the AP.
Let m=middle number, then since sum=21, m=21/3=7
Let d=common difference.
Sum of squares
(7-d)^2+7^2+(7+d)^2=155
Expand left-hand side
3*7^2-2d^2=155
d^2=(155-147)/2=4
d=+2 or -2
=+2 (common difference is positive)
Therefore the three numbers of the AP are
{7-2,7,7+2}, or
{5,7,9}
Answer:
Step-by-step explanation:140
Answer:
First on is 0.75
Second is 22/52 (not sure about this one though)
Step-by-step explanation:
For the first one:
P(aUb) = p(a) + p(b) - p(a^b)
0.6 = 0.4 + 0.5 - p(a^b)
p(a^b) = 0.3
P(a/b)= p(a^b)/p(a)
P(a/b)= 0.3/0.4
P(a/b)= 0.75
For the second one:
Total of the cards are 52
Diamonds are 13 out of that 52
Face cards are 12 out of that 52
Diamond OR face cards means p(aUb)
And that equals p(a)+p(b)-p(a^b)
p(a^b) means the intersection between the two, there are three common cards between diamonds and face cards, so p(a^b)=3
13/52 + 12/52 - 3/52= 22/52
