Ask question

Ask question

All categories

3 years ago

5

Of the 12 temporary employees in a certain company, 4 will be hired as permanent employees. If 5 of the 12 temporary employees a

re women, how many of the possible groups of 4 employees consist of 3 women and one man?A. 22B. 35C. 56D. 70E. 105

1 answer:

Aliun [14]3 years ago

3 0

Answer:

70 groups

Step-by-step explanation:

$5c_{3}*7c_{1}$

$=\frac{5*4*3}{3*2*1} *\frac{7}{1} \\=70$

You might be interested in

Hey could someone please help

Elena-2011 [213]

The anwser is4=6

the anwser is 5=26

the anwer is 6=8

8 0

3 years ago

a rectangle has a perimeter of 88cm. if the height is three times the length of the width what are the dimensions of the rectang

fredd [130]

Answer:

P=88. L=3w

P= 2(w)+2(L)

Then substitute the values into the equation

88=2w+2(3w)

Then solve for w and its length which is 3w

5 0

3 years ago

Help me please ?? ASAP!!

nirvana33 [79]

Answer:

my best guess is answer choice c.

5 0

2 years ago

Read 2 more answers

Suppose you deposited $900 into an account that earns 4% simple annual interest. Find the balance of the account after 6 months

Jet001 [13]

Balance after 6 months would be 900*(900+2%). You just need to evaluate it then you will get the answer. btw 2% comes from 4% because it's half of the year.

3 0

3 years ago

A poker hand consisting of 9 cards is dealt from a standard deck of 52 cards. Find the probability that the hand contains exactl

Andrew [12]

Answer: $\dfrac{^{12}C_{6}\times^{40}C_3}{^{52}C_9}$ or $\dfrac{228}{91885}$ .

Step-by-step explanation:

Given : A poker hand consisting of 9 cards is dealt from a standard deck of 52 cards.

The total number of cards in a deck 52

Number of faces cards in a deck = 12

Number of cards not face cards = 40

The total number of combinations of drawing 9 cards out of 52 cards = $^{52}C_9$

Now , the combination of 9 cards such that exactly 6 of them are face cards = $^{12}C_{6}\times^{40}C_3$

Now , the probability that the hand contains exactly 6 face cards will be :-

$\dfrac{^{12}C_{6}\times^{40}C_3}{^{52}C_9}$

$=\dfrac{\dfrac{12!}{6!6!}\times\dfrac{40!}{3!37!}}{\dfrac{52!}{9!\times43!}}\ \ [\because\ ^nC_r=\dfrac{n!}{r!(n-r)!}]\\\\=\dfrac{228}{91885}$

Hence, the probability that the hand contains exactly 6 face cards. is $\dfrac{228}{91885}$ .

6 0

3 years ago