Question about Solve an algebraic equation
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Details: Your company is considering various locations for expansion. It is your job to check out the various areas and the friendliness for business. One of the factors of interest is the average temperature.
As you start your drive, the first electronic sign you see states that the temperature is 22.0 °C. You are not very familiar with this temperature scale. As you drive 10 miles east, the temperature is now listed as 69.8 °F. Is this warmer or colder? The next temperature, out another 10 miles, is 20.0 °C. It seems that temperature is declining with distance traveled to the east. However, you cannot tell, unless you are able to convert from one temperature scale to another.
To do so, you will need to solve a linear equation. The linear equation that relates °C to °F is as follows:
°F = 1.8 °C + 32.0
The following data was calculated in this way:
Temperature °F Miles east from starting point
71.6 0
69.8 10
68.0 20
For the discussion, please do the following:
Provide a plot of temperature versus distance east. Based on its appearance, is it a good linear model for predicting temperature as a function of distance?
Use the graph to determine the expected temperature at 50 miles east.
If the temperature is 65 °F, determine how many miles east you are.
Discuss several other questions that can be answered using the plot.
Solve the equation for °C.
What is 75 °F in Celsius?
What is 65 °F in Celsius?
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Solution Summary
This solution consists of a detailed explanation of how to solve for an algebraic equation.
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2-4 paragraphs plus a graph
Details: Your company is considering various locations for expansion. It is your job to check out the various areas and the friendliness for business. One of the factors of interest is the average temperature.
As you start your drive, the first electronic sign you see states that ...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
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- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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