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2 years ago

I’m not very good at math so please help

Mathematics

1 answer:

Slav-nsk [51]2 years ago

4 0

Answer:

the graph is increasing from (-∞,0) and decreasing from (0,∞)

Step-by-step explanation:

By looking at the graph, the graph is increasing from (-∞,0) and decreasing from (0,∞)

The interval from (0,∞) will continue to decrease

You might be interested in

What is a positive angle less than 360 degrees that is conterminal with 540 degrees

jekas [21]

Answer:

Please check the explanation.

Step-by-step explanation:

As we have to determine a positive angle less than 360 degrees that is conterminal with 540 degrees.

Subtract 360° from 540°

540° - 360° = 180°

The resulting angle of 180° is positive, less than 360°, and coterminal with 540°.

7 0

2 years ago

What is the value of X

Volgvan

Answer:

Step-by-step explanation:

it's another one of those similar triangles

it's easier to see this one.

t /c = u / b

where

t = 7x + 3

c = 27

u = 35

b = 21

plug in the knowns

7x + 3 / 27 = 35 / 21

7x + 3 = 27 ( 35 / 21 )

7x + 3 = 9 ( 35 / 7 ) ( divided by 3 )

7x + 3 = 9 ( 5) ( 7 into 35 =5 )

7x + 3 = 45

7x = 45 - 3

7x = 42

x = 6

I checked it, the ratio is 1_2/3 :) for both, but you can check it too

7 0

3 years ago

A minor-league hockey team had the following scores for the start of the season: 3, 6, 2, 1, 2, 3, 0, 4, 5, 1, 5, 4.

ale4655 [162]

If the hockey team scores 9 goals in their next game, it means that; the mean would be the most affected.

<h3>How to find mean, median and mode?</h3>

We are given the data set as;

3, 6, 2, 1, 2, 3, 0, 4, 5, 1, 5, 4.

Let us rearrange in ascending order to get;

0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6.

The mean = (0 + 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6)/12 = 3

Median = (3 + 3)/2 = 3

Mode = 1, 2, 3, 3, 5

Now, if they score 9 goals in their next game, it means that;

Mean = 45/13 = 3.46

Median = 3

Mode remains the same

Thus, mean will be the most affected.

Read more about Mean, Median and Mode at; brainly.com/question/14532771

#SPJ1

4 0

1 year ago

The price of Veronica’s meal before tax and tip was $11.92. Veronica paid 8% tax, then added a 15% tip to the total. To the near

Alecsey [184]

Answer:

$14.80

Step-by-step explanation:

The price of Veronica's meal before tax and tip was $11.92. Veronica paid 8% tax, then added a 15% tip to the total

8 0

3 years ago

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t

skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that $\mu = 273, \sigma = 100$

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 30, s = \frac{100}{\sqrt{30}}$

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}$

$Z = 0.88$

$Z = 0.88$ has a p-value of 0.8106

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 50, s = \frac{100}{\sqrt{50}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}$

$Z = 1.13$

$Z = 1.13$ has a p-value of 0.8708

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}$

$Z = -1.13$

$Z = -1.13$ has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 100, s = \frac{100}{\sqrt{100}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}$

$Z = -1.6$

$Z = -1.6$ has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

6 0

2 years ago