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vazorg [7]
3 years ago
10

Exercise topic: Permutations and Combinations. A company wants to hire 3 new employees, but there are 8 candidates, 6 of them wh

ich are men and 2 are women. If the selection is random: a) In how many different ways can choose new employees? b) In how many different ways can choose a single male candidate? c) In how many different ways can choose at least one male candidate? with procedures. Help me please..
Mathematics
1 answer:
SOVA2 [1]3 years ago
8 0

Answer:

(a) 56 ways

(b) 6 ways

(c) 56 ways

Step-by-step explanation:

Given:

candidates: 6 mail, 1 female

number to hire : 3

a) In how many different ways can choose new employees?

use the combination formula to choose r to hire out of n candidates

C(n,r) = C(8,3) = 8! / (3! (8-3)! ) = 40320 / (120*6) = 56 ways

b) In how many different ways can choose a single male candidate?

6 ways to choose a male, one way to choose two female, so 6*1 = 6 ways

c) In how many different ways can choose at least one male candidate?

To choose at least 1 male candidate, we subtract the ways to choose no male candidates out of 56.

Since there are only two females, there is no way to choose 3 female candidates.

In other words, there are 56-0 = 56 ways (as in part (a) ) to hire 3 employees with at least one male candidate.

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9/11 = c/55<br><br><br> c=???<br> answwedbhksjbdew<br> \
Scorpion4ik [409]

Answer:

C=45

if 9/11 = c/55 you have to divide the c/55 by 9/11, resulting in c/9 = 5

multiply 9 by 5 to get 45

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
If possible help me in this. I need to find the two odd numbers...​
Lelechka [254]

Part (a)

Consecutive odd integers are integers that odd and they follow one right after another. If x is odd, then x+2 is the next odd integer

For example, if x = 7, then x+2 = 9 is right after.

<h3>Answer:  x+2</h3>

========================================================

Part (b)

The consecutive odd integers we're dealing with are x and x+2.

Their squares are x^2 and (x+2)^2, and these squares add to 394.

<h3>Answer: x^2 + (x+2)^2 = 394</h3>

========================================================

Part (c)

We'll solve the equation we just set up.

x^2 + (x+2)^2 = 394

x^2 + x^2 + 4x + 4 = 394

2x^2+4x+4-394 = 0

2x^2+4x-390 = 0

2(x^2 + 2x - 195) = 0

x^2 + 2x - 195 = 0

You could factor this, but the quadratic formula avoids trial and error.

Use a = 1, b = 2, c = -195 in the quadratic formula.

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(2)\pm\sqrt{(2)^2-4(1)(-195)}}{2(1)}\\\\x = \frac{-2\pm\sqrt{784}}{2}\\\\x = \frac{-2\pm28}{2}\\\\x = \frac{-2+28}{2} \ \text{ or } \ x = \frac{-2-28}{2}\\\\x = \frac{26}{2} \ \text{ or } \ x = \frac{-30}{2}\\\\x = 13 \ \text{ or } \ x = -15\\\\

If x = 13, then x+2 = 13+2 = 15

Then note how x^2 + (x+2)^2 = 13^2 + 15^2 = 169 + 225 = 394

Or we could have x = -15 which leads to x+2 = -15+2 = -13

So, x^2 + (x+2)^2 = (-15)^2 + (-13)^2 = 225 + 169 = 394

We get the same thing either way.

<h3>Answer: Either 13, 15  or  -15, -13</h3>
4 0
3 years ago
Which graph represents the solution to the system of equations below?
blsea [12.9K]
C. -4,-5 hope this helps :)
4 0
3 years ago
Read 2 more answers
What is the answer for this 8k−5(−5k+3)
Helen [10]

Answer:

33k-15

Step-by-step explanation:

8k−5(−5k+3)

8k+25k-15 because -5(-5) equals=25. Two negatives equal a positive.

-5(3)= -15

So the equation changes to 8k+25k-15.

Combine like terms, and you get:

33k-15

Hope this helps, have a good day. c;

6 0
3 years ago
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Can someone plz help me it is my homework or at least can someone try.Because my teacher will mark it a incomplete if it is blan
UkoKoshka [18]
Which question do you need help on ?
8 0
3 years ago
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