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2 years ago

Can someone plz help me on mixed numbers plz

Mathematics

2 answers:

IRINA_888 [86]2 years ago

3 0

Answer:

The answer 3/4

Step-by-step explanation:

olga2289 [7]2 years ago

3 0

Answer:

\frac{3}{4}

[tex]convert \: the \: mixed \: numbers \: into \: a \: fraction frac{9}{4} - \frac{3}{2}

applying the fractions formula for subtraction

(9 \times 2) - (3 \times 4) \\ 4 \times 2 \\ = 18 - 12 \\ 8 \\ = \frac{6}{8} \\ then \: diide \: each \: number \: by \: 2 \: and \: get \\ \: \frac{3}{4}

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Answer:

(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.

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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