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Flauer [41]
2 years ago
15

An academic advisor calculated the mean GPA for college freshmen to be 2.33, with a standard deviation of 1.06. The advisor know

s that there is a relationship between high school GPA and freshman-year GPA. If the advisor uses each student's high school GPA to predict his/her freshman-year GPA, what happens to the average error in prediction (relative to not using high school GPA to predict freshman-year GPa.
Mathematics
1 answer:
AfilCa [17]2 years ago
4 0

Answer:Average error will be less

Step-by-step explanation: Prediction error refers to the difference between the the actual and predicted value of an observation. Using correlation analysis or study, The level or degree of correlation or relationship between two measured variables usually determines the variation or measure of the prediction error. If there is no relationship or correlation between two measured variables, if such model is used to make prediction, the average prediction error will be very high. However, since it is stated that there exist a relationship between high school GPA and Freshman-year GPA, then making prediction based on this assertion will lessen the average prediction error.

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Multiply (400 + 20 + 3) x 10000

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423 x 10,000 = Adding the amount of zero’s in 10000 to 423



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Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
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Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
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This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
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