Help me with this i dont really get it

Please help. due today.

guapka [62]

Answer:

1, 1/2, 1 1/2

Step-by-step explanation:

Question 1

20 divided by 18 is 1.1

So the answer is closest to 1

Question 2

9 divided by 20 is 0.45

So the answer is closest to 1/2

Question 3

7 divided by 5 is 1.4

So the answer is closest to 1 1/2

Hope this helps!

3 0

3 years ago

Please help me with this question, image attached.

Brilliant_brown [7]

Answer:

AEG - BEA = BEG => 81-27=54

Step-by-step explanation:

3 0

3 years ago

Suppose a basketball player has made 231 out of 361 free throws. If the player makes the next 2 free throws, I will pay you $31.

statuscvo [17]

Answer:

The expected value of the proposition is of -0.29.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either the player will make it, or he will miss it. The probability of a player making a free throw is independent of any other free throw, which means that the binomial probability distribution is used 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.

Suppose a basketball player has made 231 out of 361 free throws.

This means that $p = \frac{231}{361} = 0.6399$

Probability of the player making the next 2 free throws:

This is P(X = 2) when n = 2. So

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

$P(X = 2) = C_{2,2}.(0.6399)^{2}.(0.3601)^{0} = 0.4095$

Find the expected value of the proposition:

0.4095 probability of you paying $31(losing $31), which is when the player makes the next 2 free throws.

1 - 0.4059 = 0.5905 probability of you being paid $21(earning $21), which is when the player does not make the next 2 free throws. So

$E = -31*0.4095 + 21*0.5905 = -0.29$

The expected value of the proposition is of -0.29.

3 0

2 years ago

A private and a public university are located in the same city. For the private university, 1038 alumni were surveyed and 647 sa

Snezhnost [94]

Answer:

The difference in the sample proportions is not statistically significant at 0.05 significance level.

Step-by-step explanation:

Significance level is missing, it is α=0.05

Let p(public) be the proportion of alumni of the public university who attended at least one class reunion

p(private) be the proportion of alumni of the private university who attended at least one class reunion

Hypotheses are:

$H_{0}$ : p(public) = p(private)

$H_{a}$ : p(public) ≠ p(private)

The formula for the test statistic is given as:

z= $\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of public university students who attended at least one class reunion ( $\frac{808}{1311}=0.616$ )

p2 is the sample proportion of private university students who attended at least one class reunion ( $\frac{647}{1038}=0.623$ )

p is the pool proportion of p1 and p2 ( $\frac{808+647}{1311+1038}=0.619$ )

n1 is the sample size of the alumni from public university (1311)

n2 is the sample size of the students from private university (1038)

Then z= $\frac{0.616-0.623}{\sqrt{{0.619*0.381*(\frac{1}{1311} +\frac{1}{1038}) }}}$ =-0.207

Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.

6 0

3 years ago

Select all the correct answers. Which expressions are equal to 10^5?

Vlada [557]

Answer:

b, d, and E

Step-by-step explanation:

when you multiplicate all, is the same results

7 0

3 years ago