You can find 50% of any total by multiplying the total by all the following except which?

Suppose that for a recent admissions class, an Ivy League college received 2,851 applications for early admission. Of this group

stepan [7]

Answer:

A) The probability is a measure of the likelihood than an event is going to occur. In this case we can calculate the probability of X as:

P(X) = X/N where X is the number of ways X occurs and N is the total number of events.

P(E) = E/N = 1033/2851 = 0.3623

P(R) = R/N = 854/2851 = 0.2995

P(D) = D/N = 964/2851 = 0.3381

B)

E and D are mutually exclusive because the students that were admitted early where not deffered to regular admission pool. There is no chance that there is a student that were admitted erarly and deffered at the same time, so the intersection between E and D (E∩D) is 0.

C) The number of students that were early accepted is 1033 and the total number of students accepted is 2375. The probability is going to be:

P = 1033/2375 = 0.4349

D)

In this case we can reformulate the question as: What is the probability of being accepted if you apply for early admission? This

Since 18% of the students deffered to regular admission pool 0.18 × 964 = 174 were admitted at the end.

The probability of being deffered and then accepted is going to be:

P(DA) = 174/2831 = 0.0610

The probability of ramdomly selecting a students that has been early accepted or deffered and then accepted is going to be:

P(E or DA) = 0.0610 + 0.3623 = 0.4233

This is the addition rule.

3 0

2 years ago

Solve the equation using square roots. 6x2 = 54

Mars2501 [29]

6x^2 = 54

6(x^2) = 54

6(x^2)/6 = 54/6

x^2 = 9

rt(x^2) = rt(9)

x = 3

hope this helps

5 0

2 years ago

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At a certain airport, 75% of the flights arrive on time. A sample of 10 flights is studied. Assume each flight is independent of

mars1129 [50]

Answer:

52.56% probability that eight or more of the flights will arrive on time.

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either it is on time, or it is not. The probability of a flight being on time is independent from other flights. So 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.

At a certain airport, 75% of the flights arrive on time.

This means that $p = 0.75$

A sample of 10 flights is studied.

This means that $n = 10$

Find the probability that eight or more of the flights will arrive on time.

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)$

In which

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

$P(X = 8) = C_{10,8}.(0.75)^{8}.(0.25)^{2} = 0.2816$

$P(X = 9) = C_{10,9}.(0.75)^{9}.(0.25)^{1} = 0.1877$

$P(X = 10) = C_{10,10}.(0.75)^{10}.(0.25)^{0} = 0.0563$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.2816 + 0.1877 + 0.0563 = 0.5256$

52.56% probability that eight or more of the flights will arrive on time.

4 0

2 years ago

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Suppose students' ages follow a skewed right distribution with a mean of 24 years old and a standard deviation of 3 years. If we

natta225 [31]

Answer: Option 'c' is correct.

Step-by-step explanation:

Since we have given that

Mean of students' age = 24 years

Standard deviation of students' age = 3 years

Sample size = number of students = 350

So, according to options,

a. The shape of the sampling distribution is approximately normal.

It is true as n >30, we will use normal.

b. The mean of the sampling distribution is approximately 24-years old.

It is true as it is given.

c. The standard deviation of the sampling distribution is equal to 5 years.

It is not true as it is given 3 years.

Hence, Option 'c' is correct.

8 0

2 years ago

A 2-column table with 10 rows. The first column is labeled x with entries negative 4, negative 3, negative 2, negative 1, 0, 1,

Murrr4er [49]

Answer- A,D,E

Step-by-step explanation:

A two column table with 5 rows. The first column, x, has the entries, negative 2, 0, 2, 4. The second column, y, has the entries, 6, 3.5, 1, negative 1.5.

Which equations represent the data in the table? Check all that apply.

y – 6 = y minus 6 equals StartFraction negative 5 Over 4 EndFraction left-parenthesis x plus 2 right-parenthesis.(x + 2)

y – 2 = –y minus 2 equals negative StartFraction 5 Over 4 EndFraction left-parenthesis x minus 1 right-parenthesis.(x – 1)

y + 2 = y plus 2 equals StartFraction negative 5 Over 4 EndFraction left-parenthesis x minus 6 right-parenthesis.(x – 6)

y – 1 = –y minus 1 equals negative StartFraction 5 Over 4 EndFraction left-parenthesis x minus 2 right-parenthesis.(x – 2)

y – 3.5 = –1.25x

Im just rephrasing the question

5 0

2 years ago

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