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levacccp [35]
3 years ago
7

WILL GIVE BRAINLEST NEED HELP ASAP

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

OPTION D

m<6 and m<3 are congruent because they are alternate interior angles.

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2500÷10

250

Therefore,The answer is 250

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Somebody please help
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4 0
4 years ago
A club has 20 women and 17 men and needs to form a committee of six members. How many committees are possible if the committee m
Scorpion4ik [409]

Answer:

2,022,456 committees

Step-by-step explanation:

From the above question, we are given the following information:

Number of women = 20

Number of men = 17

In order to form a committee of six members with at least 2 men, the number of ways we can do this is 5 ways and they are:

a) A committee of 6 men

b) A committee of 5 men and 1 woman

c) A committee of 4 men and 2 women

d) A committee of 3 men and 3 women

e) A committee of 2 men and 4 women

To solve for this we use the combination formula which is given as:

C(n, r) = nCr = n!/r!(n - r)!

Hence, the number of committees that are possible if the committee must have at least two men is calculated as

A committee of 6 men or A committee of 5 men and 1 woman or A committee of 4 men and 2 women or A committee of 3 men and 3 women or A committee of 2 men and 4 women

=[C(17, 6)]+ [C(17,5) × C(20,1)] + [C(17,4) × C(20,2)] + [C(17,3) × C(20,3)] + [C(17,2) × C(20,4)]

= [17!/6!(17 - 6)!] + [17!/5!(17 - 5)! × 20!/1!(20 - 1)!] + [17!/4!(17 - 4)! × 20!/2!(20 - 2)!] + [17!/3!(17 - 3)! × 20!/3!(20 - 3)!] + [17!/2!(17 - 2)! × 20!/4!(20 - 4)!]

= [12376] + [ 6188 × 20] + [2380 × 190] + [680 × 1140] + [ 136 ×4845]

= 2,022,456 committees

5 0
4 years ago
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