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Last weekend, 5% of the tickets sold at seaworld were discount tickets. If seaworld sold 60 tickets in all, how many dicount tic

Liula [17]

I think you have to divide 60 and 5%

5 0

3 years ago

Karen wants to place a lamp halfway between the

Bess [88]

Answer:

In D

Step-by-step explanation:

in d you can place the lamp there

3 0

2 years ago

Create triangle A’B’C by reflecting triangle ABC over the y-axis on the coordinate plane. What are the new coordinates of the ve

pickupchik [31]

Answer:

Step-by-step explanation:

When talking about reflection, we usually mean reflecting across a line, or axis. Reflecting a shape means looking at the mirror image on the other side of the axis. Consider the point (x,y), if you reflect this point across the y-axis you should multiply the x-coordinate by -1, so you get:

$(x,y)\rightarrow(-x,y)$

For triangle ABC we have:

$A(-4,2) \\ \\ B(-1,3) \\ \\ C(-1,-4)$

Therefore, the reflecting points are:

$A'(4,2) \\ \\ B(1,3) \\ \\ C(1,-4)$

Both the original triangle and the reflected one are shown below.

8 0

3 years ago

PLSSS HELLPPPPP ASAPPP

padilas [110]

I believe it would be 80

8 0

2 years ago

Read 2 more answers

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