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3 years ago

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The value of t varies directly with w. When t = 32, the value of w = 56. What is the value of w when t = 92? A 116 B 126 C 161 D

182

1 answer:

stich3 [128]3 years ago

8 0

Answer:

(D). 182

Step-by-step explanation:

did the math

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A number t is less than or equal to 5

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It would be written as
T < 5, but there would be a line underneath the less then sign

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3 years ago

You are a traffic accident investigator. You have arrived at the scene of an accident. Two trucks of equal mass (3,000 kg each)

Kazeer [188]

Answer:

1) Kinetic energy of the truck 1 is 600,000 joules.

Kinetic energy of the truck 2 is 1,837,500 joules.

2 a)Potential energy of truck 1 is 646,800 Joules.

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.

Step-by-step explanation:

Mass of the truck 1 and truck 2,m = 3000 kg

Velocity of the truck 1, $v_1$ = = 20 m./s

Velocity of the truck 2, $v_2$ = 35 m./s

1. Kinetic energy of the truck 1: $\frac{1}{2}mv_{1}^2=\frac{1}{2}\times 3000 kg\times (20 m/s)^=600,000 joules$

Kinetic energy of the truck 2: $\frac{1}{2}mv_{2}^2=\frac{1}{2}\times 3000 kg\times (35 m/s)^=1,837,500 joules$

2.a) Energy loose by the truck 1:

Initial height of the truck on the hill = 22 m

At that height potential energy will be maximum:

Potential energy of truck 1 at that height = $mgh=3000\times 9.8 m/s^2\times 22 m=646,800 Joules$

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.

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4 years ago

What is Statistical inference?

rjkz [21]

Answer:

<em>Definition 1: The theory, methods, and practice of forming judgments about the parameters of a population and the reliability of statistical relationships, typically on the basis of random sampling.</em>

<em>Definition 2: The use of randomization in sampling allows for the analysis of results using the methods of statistical inference. Statistical inference is based on the laws of probability, and allows analysts to infer conclusions about a given population based on results observed through random sampling. Two of the key terms in statistical inference are parameter and statistic.</em>

Step-by-step explanation:

Hope this helps, have a good day. c;

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3 years ago

Read 2 more answers

How is the Recursive Formula used in real life

OlgaM077 [116]

Answer:

People often sort stacks of documents using a recursive method. For example, imagine you are sorting 100 documents with names on them. First place documents into piles by the first letter, then sort each pile. Looking up words in the dictionary is often performed by a binary-search-like technique, which is recursive.

Step-by-step explanation:

This is just an example.

If this helps please mark as brainliest

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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

3 years ago