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3 years ago

12

Help me out ?<<<<<<<<<<<

1 answer:

Advocard [28]3 years ago

7 0

It would be J because the radius is 6 inches and the radius is only starting from the edge to the middle not fully across it can be confusing. Hope that helped!

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To drive to work, Dave has to drive 20 miles east and then 15 miles north. If there were a direct road going northeast, how many

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Please help me! I need this answered fast!

ArbitrLikvidat [17]

Answer:

Question 1: A

Question 2: C

Step-by-step explanation:

The answer to question 1 is A. I know this is correct because Max needs to earn 108 dollars for the headphones he wants so the amount of money he earns per hour which is 15, times the amount of hours he needs to work which is x has to be greater than 108. So, the inequality equation has to be 15x > 108 which is answer A.

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2 years ago

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Read 2 more answers

In Illinois, 9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders; that is, they have been arre

Veseljchak [2.6K]

Answer:

a) 21.58% probability that exactly 3 people are repeat offenders

b) 97.91% probability that at least one person is a repeat offender

c) 3.69

d) 1.83

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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

9% of all drivers arrested for DUI (Driving Under the Influence) are repeat offenders

This means that $p = 0.09$

41 people arrested for DUI in Illinois are selected at random.

This means that $n = 41$

a. What is the probability that exactly 3 people are repeat offenders?

This is P(X = 3).

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

$P(X = 3) = C_{41,3}.(0.09)^{3}.(0.91)^{38} = 0.2158$

21.58% probability that exactly 3 people are repeat offenders

b. What is the probability that at least one person is a repeat offender?

Either none are repeat offenders, or at least one is. The sum of the probabilities of these outcomes is 1. So

$P(X = 0) + P(X \geq 1) = 1$

We want $P(X \geq 1)$ .

Then

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = 0) = C_{41,0}.(0.09)^{0}.(0.91)^{41} = 0.0209$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0209 = 0.9791$

97.91% probability that at least one person is a repeat offender

c. What is the mean number of repeat offenders?

$E(X) = np = 41*0.09 = 3.69$

d. What is the standard deviation of the number of repeat offenders?

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{41*0.09*0.91} = 1.83$

4 0

3 years ago