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hammer [34]
3 years ago
8

Sally shares 36 pencils between 9 friends. How many pencils did each friend receive?

Mathematics
2 answers:
horsena [70]3 years ago
8 0

36 pencils ÷ 9 friends = 4 pencils per friend

Each friend received 4 pencils.

Hope this helps!! :)

Debora [2.8K]3 years ago
7 0

Answer:

4

Step-by-step explanation:

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Christine Wong has asked Dave and Mike to help her move into a new apartment on Sunday morning. She has asked them both in case
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(a) The probability that both Dave and Mike will show up is 0.25.

(b) The probability that at least one of them will show up is 0.75.

(c) The probability that neither Dave nor Mike will show up is 0.25.

Step-by-step explanation:

Denote the events as follows:

<em>D</em> = Dave will show up.

<em>M</em> =  Mike will show up.

Given:

P(D^{c})=0.55\\P(M^{c})=0.45

It is provided that the events of Dave of Mike showing up are independent of each other.

(a)

Compute the probability that both Dave and Mike will show up as follows:

P(D\cap M)=P(D)\times P (M)\\=[1-P(D^{c})]\times [1-P(M^{c})]\\=[1-0.55]\times[1-0.45]\\=0.2475\\\approx0.25

Thus, the probability that both Dave and Mike will show up is 0.25.

(b)

Compute the probability that at least one of them will show up as follows:

P (At least one of them will show up) = 1 - P (Neither will show up)

                                                   =1-P(D^{c}\cup M^{c})\\=P(D\cup M)\\=P(D)+P(M)-P(D\cap M)\\=[1-P(D^{c})]+[1-P(M^{c})]-P(D\cap M)\\=[1-0.55]+[1-0.45]-0.25\\=0.75

Thus, the probability that at least one of them will show up is 0.75.

(c)

Compute the probability that neither Dave nor Mike will show up as follows:

P(D^{c}\cup M^{c})=1-P(D\cup M)\\=1-P(D)-P(M)+P(D\cap M)\\=1-[1-P(D^{c})]-[1-P(M^{c})]+P(D\cap M)\\=1-[1-0.55]-[1-0.45]+0.25\\=0.25

Thus, the probability that neither Dave nor Mike will show up is 0.25.

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