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3 years ago

7

a street that is 116 M long is covered in snow. City workers are using a snow plow to clear the street. A tire on the snow plow

house to turn 29 x and traveling the length of the street. What is the diameter of the tire. Use the value 314 for pie

1 answer:

Anvisha [2.4K]3 years ago

8 0

Let diameter of the tire be D.
It turned 29 times and travelled 116 m in distance.

Using the circumference formula,
116=29(2 π r) = 29*(π D) = 29(3.14 D)
Solve for D
D=116/(3.14*29) = 1.274 m.

Answer: The diameter of the tire is 1.27 m (to three significant figures)

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Answer:

$\\ \mu = 118\;grams\;and\;\sigma=30\;grams$

Step-by-step explanation:

We need to use z-scores and a standard normal table to find the values that corresponds to the probabilities given, and then to solve a system of equations to find $\\ \mu\;and\;\sigma$ .

<h3>First Case: items from 100 grams to the mean</h3>

For finding probabilities that corresponds to z-scores, we are going to use here a <u>Standard Normal Table </u><u><em>for cumulative probabilities from the mean </em></u><em>(Standard normal table. Cumulative from the mean (0 to Z), 2020, in Wikipedia) </em>that is, the "probability that a statistic is between 0 (the mean) and Z".

A value of a z-score for the probability P(100<x<mean) = 22.57% = 0.2257 corresponds to a value of z-score = 0.6, that is, the value is 0.6 standard deviations from the mean. Since this value is <em>below the mean</em> ("the items produced weigh between 100 grams up to the mean"), then the z-score is negative.

Then

$\\ z = -0.6\;and\;z = \frac{x-\mu}{\sigma}$

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

<h3>Second Case: items from the mean up to 190 grams</h3>

We can apply the same procedure as before. A value of a z-score for the probability P(mean<x<190) = 49.18% = 0.4918 corresponds to a value of z-score = 2.4, which is positive since it is after the mean.

Then

$\\ z =2.4\;and\; z = \frac{x-\mu}{\sigma}$

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

<h3>Solving a system of equations for values of the mean and standard deviation</h3>

Having equations (1) and (2), we can form a system of two equations and two unknowns values:

$\\ -0.6 = \frac{100-\mu}{\sigma}$ (1)

$\\ 2.4 = \frac{190-\mu}{\sigma}$ (2)

Rearranging these two equations:

$\\ -0.6*\sigma = 100-\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

To solve this system of equations, we can multiply (1) by -1, and them sum the two resulting equation:

$\\ 0.6*\sigma = -100+\mu$ (1)

$\\ 2.4*\sigma = 190-\mu$ (2)

Summing both equations, we obtain the following equation:

$\\ 3.0*\sigma = 90$

Then

$\\ \sigma = \frac{90}{3.0} = 30$

To find the value of the mean, we need to substitute the value obtained for the standard deviation in equation (2):

$\\ 2.4*30 = 190-\mu$ (2)

$\\ 2.4*30 - 190 = -\mu$

$\\ -2.4*30 + 190 = \mu$

$\\ \mu = 118$

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Answer:

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Step-by-step explanation:

Solve : $\ln 2 + \ln x = 5$

$\ln$ represents the logarithmic function with base e.

Using logarithmic properties:

$\ln (mn) = \ln m + \ln n$

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Apply the logarithmic properties; we have

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