Answer:
<h2><em><u>Option</u></em><em><u> </u></em><em><u>B</u></em><em><u>:</u></em></h2><h3><em><u>112</u></em><em><u> </u></em><em><u>families</u></em><em><u> </u></em></h3>
Step-by-step explanation:
<em><u>Given</u></em><em><u>,</u></em>
No. of families = 75
Out of 75 no. of families had cars = 42
Out of 1, no. of families can have cars
<em><u>Therefore</u></em><em><u>, </u></em>
Out of 200 families, no. of families had cars
= 8 × 14
= 112
<em><u>Hence</u></em><em><u>, </u></em>
<em><u>option</u></em><em><u> </u></em><em><u>B</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>correct</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>112</u></em><em><u> </u></em><em><u>families</u></em><em><u> </u></em><em><u>have</u></em><em><u> </u></em><em><u>cars</u></em><em><u> </u></em><em><u>out</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>200</u></em><em><u> </u></em><em><u>families</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em>
Question:
A = {2, 3, 4, 5}
B = {4, 5, 6, 7, 8}
Two integers will be randomly selected from the sets above, one integer from set A and one integer from set B. What is the probability that the sum of the two integers will equal 9?
A. 0.15
B. 0.20
C. 0.25
D. 0.30
E. 0.33
Answer:
Option B: 0.20 is the probability of the sum of the two integers.
Explanation:
The sample space for selecting 2 numbers is given by
We need to determine the probability that the sum of two integers will be equal to 9.
Hence, we need to add the two integers from the sets A and B such that their sum will be equal to 9.
Hence, the sets are 
Thus, the total number of sets whose sum is equal to 9 = 4
The probability that the sum of the two integers will equal 9 is given by
Thus, the probability that the sum of the two integers will equal 9 is 0.20
Hence, Option B is the correct answer.
The idea of grouping is to get the terms in groups that allow you to factor out from each group something that becomes a factor in itself, leaving two factors behind that are alike. Like this: rearrange those terms so they are in this order:
(-70y^2 - 63xy) + (90x^3 + 100yx^2). Now out of the first set we are going to factor out a -7y, and out of the second set we are going to factor out a 10x^2. When we do this, this is what we get: -7y(10y + 9x) + 10x^2(9x + 10y). As you can see, what's inside both sets of parenthesis is the same, just in a different order. We can factor that out now, leaving (9x + 10y)(10x^2 - 7y). And that is factored by grouping.
Hello from MrBillDoesMath!
Answer:
I think your formula (2y/x2) is equivalent to
(2*y)/(x^2) where x^2 means x-squared.
If so, then
(2*y)/(x^2) = (2*3)/(4*4) = 6/16 = 3/8
Thank you,
MrB
