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Darina [25.2K]
3 years ago
7

Based on the line of best fit, how many students wear a jacket to school when the temperature is 80 degrees F

Mathematics
1 answer:
Snezhnost [94]3 years ago
7 0

Answer:

60 children

Step-by-step explanation:

Using the given regression graph is o make prediction on how any student wear a jacket to school at temperature = 80° F

All we have to do is locate 80 on the x axis and trave up to the point where it intersects the regression line. This intersection point is the traced to the y axis and the reading is taken, this is the prediction on the number of children who wears jacket when temperature is 80°.

By doing this, the value obtained is 60

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The point C(2,3) is translated along a vector that is parallel to the line y=3x-2. The translation vector has magnitude square r
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3 years ago
1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the fie
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Part A

Answers:
Mean = 5.7
Standard Deviation = 0.046

-----------------------

The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32
So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

================================================

Part B

The 95% confidence interval is roughly (3.73, 7.67)
The margin of error expression is z*s/sqrt(n)
The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

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ME = z*s/sqrt(n)
ME = 1.96*5.7/sqrt(32)
ME = 1.974949
The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:
L = xbar-ME
L = 5.7-1.974949
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L = 3.73
and
U = xbar+ME
U = 5.7+1.974949
U = 7.674949
U = 7.67

================================================

Part C

Confidence interval is (5.99, 6.21)
Margin of Error expression is z*s/sqrt(n)
If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*0.34/sqrt(34)
ME = 0.114286657
The margin of error is roughly 0.114286657

L = lower limit
L = xbar-ME
L = 6.1-0.114286657
L = 5.985713343
L = 5.99

U = upper limit
U = xbar+ME
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U = 6.214286657
U = 6.21
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