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2 years ago

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Last year 950 people attended a town’s annual parade. This year 1,520 people attended. What was the percent increase in attendan

ce from last year to this year?

1 answer:

juin [17]2 years ago

4 0

Answer: 60% increase

Step-by-step explanation:

First, find how many more people attended this year: 1,520 - 950 = 570.

Now, find what percent of 950 570 is: 950 * x = 570 → x = .6 * 100 = 60% :)

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11t - 6t = 45 solve the equation and enter the value of t below

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Answer:

$t = 9$

Step-by-step explanation:

$11t - 6t = 45$

$5t = 45$

<h3> $t = \frac{45}{5}$ </h3>

$t = 9$

<h3>Hope it is helpful....</h3>

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An entrepreneur faces many bureaucratic and legal hurdles when starting a new business. The World Bank collects information abou

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Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

Rearranged data:

5, 5, 5, 5, 6, 7, 7, 7, 8, 12, 12, 13, 13, 15, 18, 18, 27, 28, 36, 48, 52, 60, 66, 94

Start time for Suriname = 694

Mean = ΣX / n

Mean = (567 + 694) / (24 + 1) = 1261 / 25 = 50.44

B.) median start Time for initial 24 values

0.5(n+1)th term

0.5(25) = 12.5th term

(13 + 13) / 2 = 13

C.) preferred measure of center for the distribution will be the median as it shapes well in the middle of the distribution. The mean is overestimated

d) Compute the quartiles (Q1, Q2, Q3), and find the IQR for these data. Are there outliers in the time to start a business data set? If so, identify and name any outliers.

Using calculator :

Lower quartile Q1 --> 7

Median Q2 --> 13

Upper quartile Q3 --> 42

IQR = (Q3 - Q1) = 42 - 7 = 35

OUTLIER:

Lower bound : Q1 - 1.5(IQR) ; 7 - 1.5(35) = - 45.5

Upper bound : Q3 + 1.5(IQR) = 42 + 1.5(35) = 94.5

Outlier : values below - 45.5 and above 94.5

Hence, Surimanes start time is the only Outlier.

e)

Standard deviation for 24 countries :

Standard deviation = sqrt[Σ(X - mean)^2 / (N - 1)]

Usibg calculator :

Standard deviation for the 24 countries is 23.83

f)

Standard deviation

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2 years ago

A store had a sale where you could get 7 bags of cashews for $2. If you wanted to buy 70 bags, how much would it cost?

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Answer:

$20

Step-by-step explanation:

7 x __=70

2 x ____=___

7 x 10 = 70

2 x 10

=20

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2 years ago

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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.

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2 years ago

When you form general ideas and rules based on your experiences and observations you call that form of reasoning

lutik1710 [3]

You could probably call it INDUCTION

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2 years ago

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