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Svetach [21]
1 year ago
12

There is 4/6 of a pizza left from dinner . Each person gets 1/6 of a pizza for lunch the next day . How many people eat the pizz

a for lunch ?​
Mathematics
1 answer:
Goryan [66]1 year ago
8 0

The number of people which gets 1/6 of the pizza as required for lunch the next day is; 4 persons.

<h3>What is the number of people who get 1/6 of the pizza for lunch?</h3>

It follows from the task content that only 4/6 of the pizza is left and each individual needs 1/6 of the pizza for lunch.

On this note, the 4/6 fraction of pizza left mean: 4 × 1/6.

Therefore, only four persons get lunch.

Read more on fractions;

brainly.com/question/17220365

#SPJ1

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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.

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Step-by-step explanation:

Denote the events as follows:

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<em>M</em> =  Mike will show up.

Given:

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(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.

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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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