Any of the 11 man can be chosen, and combined with any of the 8 women.
Assume we select man1. The selected committee can be:
(m1,w1), (m1,w2), (m1,w3), (m1,w4), (m1,w5), (m1,w6), (m1,w7), (m1,w8),
so there are 8 committees selections with man1 in them.
we could repeat the same procedure for the remaining 10 men, and get 8 committees where each of them is a member.
so there are 11*8=88 ways of choosing 1 man and 1 woman.
Answer: 88
Answer:
Yes
Step-by-step explanation:
All u got to do is 26*1 and add the zero
so 26*1=26 add the zero an u get 260
Answer: 8
Step-by-step explanation: To find the least common multiple or (lcm) of 4 and 8, let's begin by listing the first few multiples of each number.
<em><u>Multiples of 4</u></em>
1 × 4 ≈ 4
2 × 4 ≈ 8
3 × 4 ≈ 12
4 × 4 ≈ 16
Notice that we skipped 0 × 6 in our list of multiples and this is because 0 × 4 is equal to 0 and our least common multiple can't be 0.
Now, let's list the multiples of 8. When listing the multiples of 8, it's a good idea to keep an eye on the list of multiples for 4 so that we will notice when we find a least common multiple.
<u><em>Multiples for 8</em></u>
1 × 8 = 8 ← is a multiple of 4
Notice that we can stop here because all other multiples that we find will be greater than 8. Therefore, the least common multiple or (lcm) of 4 and 8 is 8.
15 + 8(7)=x?
15 + 8(x)= x?
no i think it will be the first one!
15 + 8(7)= x
hope that helped?!!!
:)
Answer:
y=7
Step-by-step explanation:
62y=434
Divide both sides my 62
y=434/62
y=7
