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3 years ago

14)

Mathematics

1 answer:

Greeley [361]3 years ago

6 0

Answer:

A = 2(11 + 6x)

Step-by-step explanation:

GCF = Greatest Common Factor (remember example: 12 and 16 gcf = 4 because it is the largest number they have together. 12 = 4 * 3 and 16 = 4*4)

22 and 12's GCF is 2:

22 = 2*11

12 = 2*6

Using the distributive property, we can find out that A is the right answer.

However, let us check the other options:

B does not make sense because 6*5 is 30. 6 only applies to 12

C is not correct because it does not apply to 22 or 12

And D only applies to 12 as well because 4*3 = 12 but 4*6 is 24

A should be the answer so I hope this helps!

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3 years ago

15 players for a softball team show up for a game: (a) How many ways are there to choose 10 players to take the field? 3003 (b)

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Answer:

(a) 3003 ways

(b) 10897286400 ways

Step-by-step explanation:

Given

$n = 15$ --- 15 players

Solving (a) Ways of selecting 10 players.

This implies combination.

So, we have:

$r=10$

Using:

$^nC_r = \frac{n!}{(n-r)!r!}$

We have:

$^{15}C_{10} = \frac{15!}{(15-10)!10!}$

$^{15}C_{10} = \frac{15!}{5!10!}$

Simplify

$^{15}C_{10} = \frac{15*14*13*12*11*10!}{5!10!}$

$^{15}C_{10} = \frac{15*14*13*12*11}{5!}$

$^{15}C_{10} = \frac{15*14*13*12*11}{5*4*3*2*1}$

$^{15}C_{10} = \frac{360360}{120}$

$^{15}C_{10} = 3003$

Solving (b) Ways of assigning positions to 10 players.

This implies permutation.

So, we have:

$r=10$

Using:

$^nP_r = \frac{n!}{(n-r)!}$

We have:

$^{15}P_{10} = \frac{15!}{(15-10)!}$

$^{15}P_{10} = \frac{15!}{5!}$

Solve each factorial

$^{15}P_{10} = \frac{1307674368000}{120}$

$^{15}P_{10} = 10897286400$

Solving (c) Ways of choosing at least 1 woman

We have:

$Men = 10$

$Women = 5$

Ways of selecting 10 players is: (a) 3003 ways

Since the number of men are 10, there is 1 way of selecting 10 men (i.e. selection without women)

Using complement rule:

At least 1 woman = Total - No woman

$At\ least\ 1\ woman = 3003 - 1$

$At\ least\ 1\ woman = 3002$

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3 years ago