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saw5 [17]
4 years ago
13

Consider a family with 4 children. Assume the probability that one child is a boy is 0.5 and the probability that one child is a

girl is also 0.5, and that the events "boy" and "girl" are independent.
(a) List the equally likely events for the gender of the 4 children, from oldest to youngest. (Let M represent a boy (male) and F represent a girl (female). Select all that apply.) MMFF, FFFF, MMMF, two M's two F's, MFFF, FMMM, FFMF, FMFF, three M's one F, FFFM, MFFM, MFMF, one M three F's, FMFM, FMMF, MMFM, MMMM, FFMM, MFMM

(b) What is the probability that all 4 children are male? (Enter your answer as a fraction.) Incorrect: Your answer is incorrect. Notice that the complement of the event "all four children are male" is "at least one of the children is female." Use this information to compute the probability that at least one child is female. (Enter your answer as a fraction.)
Mathematics
1 answer:
jarptica [38.1K]4 years ago
5 0

Answer:

a) Total 16 possibilities

MMMM

FFFF

MMMF

MMFM

MFMM

FMMM

FFFM

FFMF

FMFF

MFFF

MMFF

MFMF

MFFM

FFMM

FMMF

FMFM

b) P(MMMM) = 1/16

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It is believed that the average starting salary for a 21-25 year old college grad exceeds $52,000 per year. Hence, it is desired
gregori [183]

Answer:

The significance level α is 0.02.

Step-by-step explanation:

A hypothesis test for single mean can be performed to determine whether the average starting salary for a 21-25 year old college grad exceeds $52,000 per year.

The hypothesis is defined as follows:

<em>H₀</em>: The average starting salary for a 21-25 year old college grad does not exceeds $52,000 per year, i.e. <em>µ</em> ≤ 52,000.

<em>Hₐ</em>: The average starting salary for a 21-25 year old college grad exceeds $52,000 per year, i.e. <em>µ</em> > 52,000.

The information provided is:

<em>σ</em> = $5,745

<em>n</em> = 65

Also, if \bar X>\$53,460 then the null hypothesis will be rejected.

Here, we need to compute the value of significance level <em>α</em>, the type I error probability.

A type I error occurs when we reject a true null hypothesis (H<em>₀</em>).

That is:

<em>α</em> = P (type I error)

<em>α</em> = P (Rejecting H<em>₀</em>| H<em>₀</em> is true)

   =P(\bar X>53460|\mu \leq 52000)

   =P[\frac{\bar X-\mu_{0}}{\sigma/\sqrt{n}}>\frac{53460-52000}{5745/\sqrt{65}}]

   =P(Z>2.05)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the significance level α is 0.02.

5 0
3 years ago
Write the product as a trinomial.
Andru [333]
First we multiply 2r*r and it gives us 2r^{2}. 

Then -5r, + 20r = 15r

Then -5*10 gives us -50.

So the answer is: 2r^{2}+15r-50
3 0
3 years ago
Read 2 more answers
A publisher reports that 75% of their readers own a particular make of car. A marketing executive wants to test the claim that t
Korolek [52]

Answer:

The p-value of the test is of 0.0536 > 0.02, which means that there is not sufficient evidence at the 0.02 level to support the executive's claim.

Step-by-step explanation:

A publisher reports that 75% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage.

At the null hypothesis, we test if the proportion is of 75%, that is:

H_0: p = 0.75

At the alternate hypothesis, we test if the proportion is different of 75%, that is:

H_1: p \neq 0.75

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

75% is tested at the null hypothesis:

This means that \mu = 0.75, \sigma = \sqrt{0.75*0.25}

A random sample of 280 found that 70% of the readers owned a particular make of car.

This means that n = 280, X = 0.7

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.7 - 0.75}{\frac{\sqrt{0.75*0.25}}{\sqrt{280}}}

z = -1.93

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion differing from 0.75 by at least |0.7 - 0.75| = 0.05, which is P(|z| > 1.93), which is 2 multiplied by the p-value of z = -1.93.

Looking at the z-table, z = -1.93 has a p-value of 0.0268.

2*0.0268 = 0.0536.

The p-value of the test is of 0.0536 > 0.02, which means that there is not sufficient evidence at the 0.02 level to support the executive's claim.

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3 years ago
For homework, Brooke has 15 math problems, 5 social studies problems, and 9 science problems. Use mental math to determine how m
koban [17]

Answer:

29 homework problems

Addition Property was used

Step-by-step explanation:

15+5=20+9=29

Addition

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4 years ago
The odds against rolling a number greater than 2<br> are
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Is that a dice your talking about?
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