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3 years ago

2 QUESTIONS 20 POINTS AND BRAINIEST IF YOU GET THEM CORRECT! THANK YOU! <3

Mathematics

2 answers:

kenny6666 [7]3 years ago

8 0

For Question 4, you first need to find how many patients went to buy contacts and how many patients went to buy glasses.

According to the graph, the number of people who bough contacts was 11 (between 10 and 12), and the number of people who bought glasses was 4.

Subtract these numbers: 11 - 4 = 7 patients, or choice (B).

For Question 2, look at the data. The lowest is 228 and the highest is 385. You would want a range that fits this best, and the range 200 to 400 would be best without cramming and without too much empty space.

The answer is (D).

kari74 [83]3 years ago

7 0

The answer to the first question is 7

Second one is 200-400

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THe halloween haunt estimated they would make $125,450 this year. They expect to make 23,700 more than last year by the time the

sveticcg [70]

Answer:

it is at least $300000 good luck

7 0

3 years ago

At a certain school,5/9 of the students are boys. Today, 2/3 of the boys brought their lunch. What fraction of the students are

Elodia [21]

5/6 of the boys brought their lunch.

In order to determine the answer you need to divide 5/9 by 2/3 in which will give you 5/6.

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3 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

3 years ago

The mean annual income for people in a certain city is 37 thousand dollars, with a standard deviation of 28 thousand dollars. A

Aloiza [94]

Answer:

$P( 31 < \bar X< 41)$

And we can ue the z score formula given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got for the limits:

$z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515$

$z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01$

So we want to find this probability:

$P(-1.515$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the annual income of a population, and for this case we know the following info:

$\mu=37$ and $\sigma=28$ and we are omitting the zeros from the thousand to simplify calculations

We select a sample size of n=50>30.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want to find this probability:

$P( 31 < \bar X< 41)$

And we can ue the z score formula given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got for the limits:

$z = \frac{31-37}{\frac{28}{\sqrt{50}}}= -1.515$

$z = \frac{41-37}{\frac{28}{\sqrt{50}}}= 1.01$

So we want to find this probability:

$P(-1.515$

4 0

3 years ago

The language arts teacher wants to know whether the students in the entire school prefer a debate or an speech. The teacher draw

Georgia [21]

The language art teacher wants to know the opinion of the ENTIRE school, so it would be fair if the teacher takes a sample from the ENTIRE school population. She can use the stratified sampling method where she can take a proportionate sample from each grade.

Correct answer: All students in each grade

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3 years ago