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const2013 [10]
3 years ago
11

According to a​ report, 51.6​% of murders are committed with a firearm. ​(a) if 200 murders are randomly​ selected, how many wou

ld we expect to be committed with a​ firearm? ​(b) would it be unusual to observe 122 murders by firearm in a random sample of 200 ​murders? why?
Mathematics
1 answer:
dolphi86 [110]3 years ago
6 0

Answer:

a) 103, b) No

Step-by-step explanation:

a) We need to multiply the probability by the amount of the sample to get:

200 × 51.6% = 103.2, rounded down to 103

b) As we have selected the people randomly, we have no control over the type of people we are given - this is theoretical and the estimate is not definite - all of the sample could have murdered by firearm or even none (even though it is highly unlikely).

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weqwewe [10]
Answer:
x= -8

Explain:
You first have to make the variable by itself so you subtract 3 on both sides. The x is by itself and you finish by subtracting -5 to 3 and you get the answer x = -8

Hope this helps
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3 years ago
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Solve –3x = 36 for x. A. x = 9 B. x = 12 C. x = –12 D. x = –9
lawyer [7]

Answer: C

Step-by-step explanation:

-3x=36

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Find the nth term<br><br> 1, 5, 9. 13, 17
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Assume that there are an equal number of births in each month so that the probability is that a person chosen at random was born
NikAS [45]

Answer:

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Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have a birthday in May, or they do not. The probability of a person having a birthday in May is independent of any other person. This means that the binomial probability distribution is used to solve this question.

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.

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And p is the probability of X happening.

Probability of a person being in May:

May has 31 days in a year of 365. So

p = \frac{31}{365} = 0.0849

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What is the probability that at the May celebration, exactly two members of the group have May birthdays?

This is P(X = 2).

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P(X = 2) = C_{20,2}.(0.0849)^{2}.(0.9151)^{18} = 0.2773

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

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Step-by-step explanation:

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