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Evgesh-ka [11]
3 years ago
13

Please answer both thank you​

Mathematics
1 answer:
jeyben [28]3 years ago
7 0

Answer:

7.

Solution given;

male=15

female=27

1st term=5*3

2nd term=3*3*3

now

Highest common factor=3

So

<u>The</u><u> </u><u>maximum</u><u> </u><u>number</u><u> </u><u>of</u><u> </u><u>groups</u><u> </u><u>that</u><u> </u><u>the</u><u> </u><u>teacher</u><u> </u><u>can</u><u> </u><u>make</u><u> </u><u>is</u><u> </u><u>3</u><u>.</u>

<u>and</u><u> </u><u>each</u><u> </u><u>team</u><u> </u><u>contains</u><u> </u><u>5</u><u> </u><u>male and</u><u> </u><u>9</u><u> </u><u>female</u><u>.</u>

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Answer:

a) P=0.0000843

b) P=0.0000454

c) P=0.0066113

Step-by-step explanation:

6 cards are drawn at random without replacement from a standard deck.

a) We calculate the probability that all the cards are hearts.

We calculatethe number of possible combinations:

C^{52}_6=\frac{52!}{6!(52-6)!}=20358520

We calculate the number of favorable combinations:

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Therefore, the probability is

P=\frac{1716}{20358520}\\\\P=0.0000843

b) We calculate the probability that all the cards are face cards.

We calculatethe number of possible combinations:

C^{52}_6=\frac{52!}{6!(52-6)!}=20358520

We calculate the number of favorable combinations:

C_6^{12}=\frac{12!}{6!(12-6)!}=924

Therefore, the probability is

P=\frac{924}{20358520}\\\\P=0.0000454

c) We calculate the probability that all the cards are even.

We calculatethe number of possible combinations:

C^{52}_6=\frac{52!}{6!(52-6)!}=20358520

We calculate the number of favorable combinations:

C_6^{24}=\frac{24!}{6!(24-6)!}=134596

Therefore, the probability is

P=\frac{134596}{20358520}\\\\P=0.0066113

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When the radius of a circle is multiplied by a factor of 2 22, by what factor will the area of the circle change?
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A = pi (r)^2
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