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3 years ago

Check my work? Rational and irrational numbers.

Mathematics

2 answers:

Sauron [17]3 years ago

3 0

Here, Rational numbers are: = 2/9 and √225
& other two are irrational

Hope this helps!

telo118 [61]3 years ago

3 0

43 is the asnwer hope it helped

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Your name is the answer to the riddle.

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Use rounding or compatible numbers to estimate the sum 790+112

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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?

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

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3 years ago

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Find the solution set of the inequality:<br><br> 8x + 2 > 34

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