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uranmaximum [27]
2 years ago
13

Can someone help me with 18 and 20 please

Mathematics
2 answers:
mihalych1998 [28]2 years ago
7 0

Answer:

both are b

Step-by-step explanation:

the point is equal to -2 because there is a point on -2 and with 20 it b because 2 time 10 is 20 +26 makes 46 which makes 43 less than 46

slamgirl [31]2 years ago
5 0
Both are b because they both show less than or equal to
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Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba
nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem, we have that:

n = 12, p = 0.5

(a) 2 or 5 children

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161

P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934

p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002

P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029

P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161

P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537

P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208

P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537

P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161

P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029

P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002

P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002

P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029

P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161

P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537

P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

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3 years ago
Which expression can be used to find 25 percent of 88?
Dominik [7]

Answer: 25/100* 88

Step-by-step explanation:

6 0
2 years ago
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An object falls from a height of 400 feet. Its height, h, in feet is given by the equation h=-16t^2+400 , where t is the time in
LekaFEV [45]
16t^2=400
t^2=25
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Help find the inequalities of these problems! Plz &amp; thank you!!
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Consider a simple example of moral hazard. Suppose that Woodrow goes into a casino to make one bet a day. The casino is very bas
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Answer:

The expected value of the safe bet equal $0

Step-by-step explanation:

If  

S=\left\{s_1,s_2,...,s_n\right\}

is a finite numeric sample space and

P(X=s_k)=p_k for k=1, 2,..., n

is its probability distribution, then the expected value of the distribution is defined as

E(X)=s_1P(X=s_1)+s_2P(X=s_2)+...+s_nP(X=s_n)X)

What is the expected value of the safe bet?

In the safe bet we have only two possible outcomes: head or tail. Woodrow wins $100 with head and “wins” $-100 with tail So the sample space of incomes in one bet is

S = {100,-100}

Since the coin is supposed to be fair,  

P(X=100)=0.5

P(X=-100)=0.5

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E(X) = 100*0.5 - 100*0.5 = 0

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3 years ago
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