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3 years ago

A graph shows the horizontal axis numbered 1 to 5 and the vertical axis numbered 1 to 5. Points and a line show an upward trend.

Mathematics

2 answers:

KIM [24]3 years ago

6 0

letter C on ed2020

i gotta write more i guess to make an answer lol

*smile*

anzhelika [568]3 years ago

5 0

Answer:

0.19 or C on ed!!!

Step-by-step explanation:

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Manny worked all summer to save money. Now he has $150 with which he wants to buy a digital music player and as many songs as he

True [87]

Answer:

Step-by-step explanation:

Total amount of Money saved by Manny is $150. He wants to buy a digital player with it.

The digital music player costs $89, and each song costs $0.99. Let x represent the number of songs that he can afford. This means that the cost of x songs would be

89 + 0.99x

If Manny spends no more than $150, before tax, the number of songs that he can buy will be

89 + 0.99x = 150

0.99x = 150 - 89 = 61

x = 61/0.99 = 61.6

He can buy 61 songs

8 0

3 years ago

4. One in four people in the US owns individual stocks. You randomly select 12 people and ask them if they own individual stocks

BartSMP [9]

Answer:

a. The mean is 3, the variance is 2.25 and the standard deviation is 1.5.

b. 0.0401 = 4.01% probability that the number of people who own individual stocks is exactly six.

c. 0.1584 = 15.84% probability that the number of people who say they own individual stocks is at least two.

d. 0.3907 = 39.07% probability that the number of people who say they own individual stocks is at most two

e. Both cases include one common outcome, that is, 2 people owning stocks, so the events are not mutually exclusive.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they own stocks, or they do not. The probability of a person owning stocks is independent of any other person, 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.

One in four people in the US owns individual stocks.

This means that $p = \frac{1}{4} = 0.25$

You randomly select 12 people and ask them if they own individual stocks.

This means that $n = 12$

a. Find the mean, variance, and standard deviation of the resulting probability distribution.

The mean of the binomial distribution is:

$E(X) = np$

$E(X) = 12(0.25) = 3$

The variance is:

$V(X) = np(1-p)$

$V(X) = 12(0.25)(0.75) = 2.25$

Standard deviation is the square root of the variance, so:

$\sqrt{V(X)} = \sqrt{2.25} = 1.5$

The mean is 3, the variance is 2.25 and the standard deviation is 1.5.

b. Find the probability that the number of people who own individual stocks is exactly six.

This is P(X = 6). So

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

$P(X = 6) = C_{12,6}.(0.25)^{6}.(0.75)^{6} = 0.0401$

0.0401 = 4.01% probability that the number of people who own individual stocks is exactly six.

c. Find probability that the number of people who say they own individual stocks is at least two.

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = C_{12,0}.(0.25)^{0}.(0.75)^{12} = 0.0317$

$P(X = 1) = C_{12,1}.(0.25)^{1}.(0.75)^{11} = 0.1267$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0317 + 0.1267 = 0.1584$

0.1584 = 15.84% probability that the number of people who say they own individual stocks is at least two.

d. Find the probability that the number of people who say they own individual stocks is at most two.

This is:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

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

$P(X = 0) = C_{12,0}.(0.25)^{0}.(0.75)^{12} = 0.0317$

$P(X = 1) = C_{12,1}.(0.25)^{1}.(0.75)^{11} = 0.1267$

$P(X = 2) = C_{12,2}.(0.25)^{2}.(0.75)^{10} = 0.2323$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0317 + 0.1267 + 0.2323 = 0.3907$

0.3907 = 39.07% probability that the number of people who say they own individual stocks is at most two.

e. Are the events in part c. and in part d. mutually exclusive

Both cases include one common outcome, that is, 2 people owning stocks, so the events are not mutually exclusive.

5 0

3 years ago

Identifying each step by step in the process<br> 4x+3=23

kramer

Answer:

x = 5

Step-by-step explanation:

4x + 3 = 23 ( subtract 3 from both sides )

4x = 20 ( divide both sides by 4 )

x = 5

6 0

3 years ago

Read 2 more answers

Help plz due today and don't worry abt my answers

Elanso [62]

This is for the one with the green background :) I corrected a few of the errors that collided with the answers for the 3 qns you didn’t know. Hope this helps!!!

8 0

3 years ago

In constructing a 95 percent confidence interval, if you increase n to 4n, the width of your confidence interval will (assuming

Damm [24]

Answer:

about 50 percent of its former width.

Step-by-step explanation:

Let's assume that our parameter of interest is given by $\theta$ and in order to construct a confidence interval we can use the following formula:

$\hat \theta \pm ME(\hat \theta)$

Where $\hat \theta$ is an estimator for the parameter of interest and the margin of error is defined usually if the distribution for the parameter is normal as:

$ME = z_{\alpha} SE$

Where $z_{\alpha/2}$ is a quantile from the normal standard distribution that accumulates $\alpha/2$ of the area on each tail of the distribution. And $SE$ represent the standard error for the parameter.

If our parameter of interest is the population proportion the standard of error is given by:

$SE= \frac{\hat p (1-\hat p)}{n}$

And if our parameter of interest is the sample mean the standard error is given by:

$SE = \frac{s}{\sqrt{n}}$

As we can see the standard error for both cases assuming that the other things remain the same are function of n the sample size and we can write this as:

$SE = f(n)$

And since the margin of error is a multiple of the standard error we have that $ME = f(n)$

Now if we find the width for a confidence interval we got this:

$Width = \hat \theta + ME(\hat \theta) -[\hat \theta -ME(\hat \theta)]$

$Width = 2 ME (\hat \theta)$

And we can express this as:

$Width =2 f(n)$

And we can define the function $f(n) = \frac{1}{\sqrt{n}}$ since as we can see the margin of error and the standard error are function of the inverse square root of n. So then we have this:

$Width_i= 2 \frac{1}{\sqrt{n}}$

The subscript i is in order to say that is with the sample size n

If we increase the sample size from n to 4n now our width is:

$Width_f = 2 \frac{1}{\sqrt{4n}} =2 \frac{1}{\sqrt{4}\sqrt{n}} =\frac{2}{2} \frac{1}{\sqrt{n}} =\frac{1}{\sqrt{n}} =\frac{1}{2} Width_i$

The subscript f is in order to say that is the width for the sample size 4n.

So then as we can see the width for the sample size of 4n is the half of the wisth for the width obtained with the sample size of n. So then the best option for this case is:

about 50 percent of its former width.

7 0

3 years ago