Please show ur work

IRINA_888 [86]

The value of 32 in the equation will serve as the time taken by the driver to travel

<h3>Functions and values</h3>

If Alice fills up the gas tank of her car before going for a long drive, and the equation that shows the amount of gas in gallons in Alice's car when she has driven m (miles). is given as:

g = 15 - m/32

The value of 32 in the equation will serve as the time taken by the driver to travel

Learn more on distance and time here; brainly.com/question/17273444

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6 0

1 year ago

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

2 years ago

I NEED HELP PLEASE. THANK YOU

vova2212 [387]

Answer:

Given

Distributive property

Subtraction property of equality

Simplify

Subtraction property of equality

Simplify

Division property of equality

Simplify

Step-by-step explanation:

If you want an explanation I would be happy to provide you with one, but I do not want to have to explain the entire thing right here because it would be really long.

7 0

2 years ago

The ratio of girls to boys in French class is 5:4. There are 12 boys in the class. How many girls are there?

jolli1 [7]

Answer:

15

Step-by-step explanation:

5/4=15/12

8 0

3 years ago

Read 2 more answers

A company studied the number of lost-time accidents occurring at its Brownsville, Texas, plant. Historical records show that 20%

zhuklara [117]

Answer:

a. 5% of the employees will experience lost-time accidents in both years

b. 24% of the employees will suffer at least one lost-time accident over the two-year period

Step-by-step explanation:

a. What percentage of the employees will experience lost-time accidents in both years?

20% last year, of those who suffered last year, 25% during this year. So

$p = 0.2*0.25 = 0.05$

5% of the employees will experience lost-time accidents in both years.

b. What percentage of the employees will suffer at least one lost-time accident over the two-year period?

5% during the two years.

10% during the current year. 25% of employees who had lost-time accidents last year will experience a lost-time accident during the current year.

So the 10% is composed of 5% during both years(25% of 20%) and 5% of the 80% who did not suffer during the first year.

First year yes, not on the second.

75% of 20%. So, total:

$0.05 + 0.05*0.8 + 0.75*0.2 = 0.24$

24% of the employees will suffer at least one lost-time accident over the two-year period

3 0

2 years ago