Find the number of terms common to the two A.P.'s 3,7,11,......,407 and 2,9,16,......,709
1 answer:
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Answer:
One possible solution be -
Mrs. Cho is brings type A cans = 10
Mrs. Cho is brings type B cans = 3
Step-by-step explanation:
Let us assume
Mrs. Cho is brings type A cans = x
Mrs. Cho is brings type B cans = y
As given,
For a type A can, she gets 5 cents and For type B can, she gets 10 cents
As we have x Type A cans and y Type B cans
And she has redeemed no more than 95 cents
⇒5x + 10y ≤ 95 .......(1)
Also given,
She has redeemed at least 11 cans
⇒ x + y ≥ 11 ........(2)
∴ we get
x + y ≥ 11
5x + 10y ≤ 95
By applying graphical method , we get
The equations become
5x + 10y = 95 .......(1)
x + y = 11 ........(2)
Table for equation (1) -
x y
0 9.5
19 0
Table for equation (2) -
x y
0 11
11 0
The two lines intersect at point (3, 8)
So the possible combinations are in the shaded portion which is covered by both the lines.
One possible combination is ( 10, 3) which lies in the shaded region.
So,
Mrs. Cho is brings type A cans = x = 10
Mrs. Cho is brings type B cans = y = 3
Answer:
a.) 0.7063
b.) 23
Step-by-step explanation:
a.)
Let X be an event in which at least 2 students have same birthday
Y be an event in which no student have same birthday.
Now,
P(X) + P(Y) = 1
⇒P(X) = 1 - P(Y)
as we know that,
Probability of no one has birthday on same day = P(Y)
⇒P(Y) = where there are n people in a group
As given,
n = 30
⇒P(Y) = = 0.2937
∴ we get
P(X) = 1 - 0.2937 = 0.7063
So,
The probability that at least two of them have their birthdays on the same day = 0.7063
b.)
Given, P(X) > 0.5
As
P(X) + P(Y) = 1
⇒P(Y) ≤ 0.5
As
P(Y) =
We use hit and trial method
If n = 1 , then
P(Y) = = 1
0.5
If n = 5 , then
P(Y) = = 0.97
0.5
If n = 10 , then
P(Y) = = 0.88
0.5
If n = 15 , then
P(Y) = = 0.75
0.5
If n = 20 , then
P(Y) = = 0.588
0.5
If n = 22 , then
P(Y) = = 0.52
0.5
If n = 23 , then
P(Y) = = 0.49
0.5
∴ we get
Number of students should be in class in order to have this probability above 0.5 = 23