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2 years ago

What is the product of 2/5 and 4/6

Mathematics

2 answers:

exis [7]2 years ago

6 0

Answer:

4/15 is the correct answer of this question.hope this helps

Aneli [31]2 years ago

4 0

Answer:

$\frac{2}{5} \times \frac{4}{6} \\ = \frac{4 \times 2}{5 \times 6} \\ = \frac{8}{30} \\ = \frac{4}{15}$

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According to government data, 75% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women

blagie [28]

Answer:

We are given that According to government data, 75% of employed women have never been married.

So, Probability of success = 0.75

So, Probability of failure = 1-0.75 = 0.25

If 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

We will use binomial

Formula : $P(X=r) =^nC_r p^r q^{n-r}$

At x = 2

$P(X=r) =^{15}C_2 (0.75)^2 (0.25^{15-2}$

$P(X=2) =\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{13}$

$P(X=2) =8.8009 \times 10^{-7}$

b. That at most 2 of them have never been married?

At most two means at x = 0 ,1 , 2

So, $P(X=r) =^{15}C_0 (0.75)^0 (0.25^{15-0}+^{15}C_1 (0.75)^1 (0.25^{15-1}+^{15}C_2 (0.75)^2 (0.25^{15-2}$

$P(X=r) =(0.75)^0 (0.25^{15-0}+15 (0.75)^1 (0.25^{15-1}+\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{15-2})$

$P(X=r) =9.9439 \times 10^{-6}$

c. That at least 13 of them have been married?

P(x=13)+P(x=14)+P(x=15)

$={15}C_{13}(0.75)^{13} (0.25^{15-13})+{15}C_{14} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})$

$=\frac{15!}{13!(15-13)!}(0.75)^{13} (0.25^{15-13})+\frac{15!}{14!(15-14)!} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})$

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3 years ago

What is the factored form of x^12x^18+1

AlladinOne [14]

I think the answer is 16x^2+1

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3 years ago

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5. Simplify: 56 = 7 - 3 [4 + {8 - 4 ( 4 + 5 - 3 )}]

shutvik [7]

Answer:

Step-by-step explanation:

56=7-3[4+{8-4(4+5-3)}]

56=7-3[4+{4(4+5-3)}]

56=7-3[4+{16+20-12)}]

56=7-3[4+24]

56=4[28]

56=92

56-56=92-56

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3 years ago

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