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Dmitry_Shevchenko [17]
2 years ago
3

A committee of three people is to be chosen from four married couples. What is the number of different committees that can be ch

osen if two people who are married to each other cannot both serve on the committee?
Mathematics
1 answer:
Grace [21]2 years ago
8 0

Answer: 32

Step-by-step explanation:

Since three (3) people are to be chosen from four (4) married couples without having two(2) married people in the same committee , it means each couple can only send one representative to the committee.

We can represent the couple using this illustration

MALE FEMALE

1. 1

2 2

3. 3

4. 4

For the first member of the committee, there are eight (8) possible choices.

(Assume we pick Female 1)

For the second member of the committee, there are six (6) possible choices. This is because we already picked Female 1and cannot pick Male 1. Let’s assume this time, we pick Male 2.

For the third member of the committee, we are left with four (4) choices. Don’t forget we already picked Female 1 and Male 2. That means we cannot pick Male 1 and Female 2.

The committee can be arranged in 3! Ways

3! = 3 × 2 × 1

= 6 ways

The different ways of selecting committee members can be gotten by multiplying the possible outcome for the first, second and third member and then divide by the ways the committee can be arranged (6)

1st member × 2nd member × 3rd member/ 6

= 32

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Alice and Briana each participate in a 5 kilometer race. Alice's distance covered, in kilometers, after t minutes can be modeled
Pavel [41]

Answer:

a. Alice

b. Briana

c. 0.51 minutes

Step-by-step explanation:

a. Alice formula is valid for any t > 0 minutes, but Briana formula is only valid for

2t - 1 > 0

2t > 1

t > 1/2 minutes

b. They finish when their covered distance is equal to 5 kilometers. For Alice:

t/4 = 5

t = 5*4 = 20 minutes

For Briana:

√(2t - 1) = 5

2t - 1 = 5²

2t = 25 + 1

t = 26/2

t = 13 minutes

c. They are side by side when they have covered the same distance, that is:

t/4 = √(2t - 1)

(t/4)² = 2t - 1

t²/16 = 2t - 1

t² = 16*(2t - 1)

t² = 32t - 16

t² - 32t + 16 = 0

Using quadratic formula:

t = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}

t = \frac{32 \pm \sqrt{-32^2 - 4(1)(16)}}{2(1)}

t = \frac{32 \pm 30.98}{2}

t_1 = \frac{32 + 30.98}{2}

t_1 = 31.49

t_2 = \frac{32 - 30.98}{2}

t_2 = 0.51

Only the second answer has sense for this problem because the race already finished before they spent 31.49 minutes in it.

4 0
2 years ago
Read 2 more answers
28. $400 is divided between Kas, Jaspar and Jae so that Kas has twice
Inga [223]

Let's define our variables before anything else:

Kas = S

Jaspar = R

Jae = E

$400 is divided between Kas, Jaspar and Jae

This means S + R + E = 400

Kas has twice  as much as Jaspar

This means that S = 2R

Jaspar has three times as much as Jae

This means that R = 3E

Because both of our smaller formulas have R, let's replace everything in the first formula with R

S + R + E = 400

plug in values

2R + R + R/3 = 400

multiply everything by 3 to get rid of the denominator

6R + 3R + R = 1200

add

10R = 1200

divide

R = 120

So Jaspar receives $120

3 0
2 years ago
Please help me with these questions and show how u got them, i will mark you brainliest !
Aneli [31]
A+7
4+7
4+7=11

b-3
5-3
5-3=2

9c
9(10)
9(10)=90
4 0
2 years ago
Drew burned 2,000 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing
tensa zangetsu [6.8K]
To answer this, write a system of equations and use one to solve the other.

b + 2c = 2000 or b = 2000 - 2c (substitute this in for b in the 2nd equation)

2b + 3c = 3300
2 (2000 - 2c) + 3c = 3300
4000 - 4c + 3c = 3300
4000 -c = 3300
-c = -700
c = 700 calories

He burned 700 calories per hour canoeing and 600 calories per hour playing basketball (2000 - 2 x 700).


 





6 0
2 years ago
Luka's weekly earnings of $475 are increased by 6% calculate his new weekly earnings
KengaRu [80]
He earns 28.5. (475•0.06)
6 0
2 years ago
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