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3 years ago

5

The table shows the total number of customers at a car wash after 1, 2, 3, and 4 days of its grand opening. which graph represen

t the data shown in the table?

1 answer:

kozerog [31]3 years ago

4 0

Answer:

Has to be the first one....

Step-by-step explanation:

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In a survey of 300 college graduates, 53% reported that they entered a profession closely related to their college major. If 9 o

ExtremeBDS [4]

Answer:

0.1348 = 13.48% probability that 3 of them entered a profession closely related to their college major.

Step-by-step explanation:

For each graduate, there are only two possible outcomes. Either they entered a profession closely related to their college major, or they did not. The probability of a graduate entering a profession closely related to their college major is independent of other graduates. This, coupled with the fact that they are chosen with replacement, means that we use the binomial probability distribution 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.

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

And p is the probability of X happening.

53% reported that they entered a profession closely related to their college major.

This means that $p = 0.53$

9 of those survey subjects are randomly selected

This means that $n = 9$

What is the probability that 3 of them entered a profession closely related to their college major?

This is P(X = 3).

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

$P(X = 9) = C_{9,3}.(0.53)^{3}.(0.47)^{6} = 0.1348$

0.1348 = 13.48% probability that 3 of them entered a profession closely related to their college major.

6 0

3 years ago

How many times greater is the value of the digit 6 in 659,451 than the value of the digit 6 in 751,632?

MrRissso [65]

It is a hundred times greater

5 0

3 years ago

Sum or products of 4x47

andrezito [222]

You multiply 4 and 47 so 188 would be the answer. Hope this helps! :)

4 0

3 years ago

Read 2 more answers

Find the circumference a circle with the diameter AB<br><br> A (-8,4) and B (4,-1)

Simora [160]

Answer:

<em>C = 13π ≈ 40.82 units </em>

Step-by-step explanation:

C = 2 r π = d π

AB = $\sqrt{(-8-4)^2 +(4+1)^2}$ = 13 units

<em>C = 13π ≈ 40.82 units</em> ( π ≈ 3.14 )

7 0

3 years ago

1. A report from the Secretary of Health and Human Services stated that 70% of single-vehicle traffic fatalities that occur at n

Nuetrik [128]

Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

For each fatality, there are only two possible outcomes, either it involved an intoxicated driver, or it did not. The probability of a fatality involving an intoxicated driver is independent of any other fatality, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

70% of fatalities involve an intoxicated driver, hence $p = 0.7$ .

A sample of 15 fatalities is taken, hence $n = 15$ .

The probability is:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)$

Hence

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

$P(X = 10) = C_{15,10}.(0.7)^{10}.(0.3)^{5} = 0.2061$

$P(X = 11) = C_{15,11}.(0.7)^{11}.(0.3)^{4} = 0.2186$

$P(X = 12) = C_{15,12}.(0.7)^{12}.(0.3)^{3} = 0.1700$

$P(X = 13) = C_{15,13}.(0.7)^{13}.(0.3)^{2} = 0.0916$

$P(X = 14) = C_{15,14}.(0.7)^{14}.(0.3)^{1} = 0.0305$

$P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047$

Then:

$P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215$

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

A similar problem is given at brainly.com/question/24863377

5 0

3 years ago