A: 2.86 B: 2.75 C: 2.95 D: 1.17
2 answers:
No doubt your text advises how to arrive at a best-fit exponential model. The two graphing calculators I used gave different answers. The first one modeled the function as
y = 398.494(0.50526)^x
For x = 7, this model gives y = 3.35
The second calculator, a TI-84 work-alike, modeled the function as
y = 422.110(0.48896)^x
For x=7, this model gives y = 2.82
Using the first and last data points only, the model is
y = 200(0.06)^((x-1)/4)
For x=7, this model gives y = 2.94
Using yet another calculator to do linear regression on the log of y, the function is modeled by
y = 10^(2.62543-0.31073x)
which is another way to write the function produced by the TI-84 work-alike.
So, the technology I used gives a couple of reasons to choose
A. 2.86
and a couple of reasons to choose
C. 2.95
Of the models shown here, the ones with the lowest mean absolute deviation and the least squared error were the ones that produced the highest values of y for x=7, suggesting 2.95 is the best choice.
It would be advisable to use the procedure described in your text.
y = 398.494(0.50526)^x
For x = 7, this model gives y = 3.35
The second calculator, a TI-84 work-alike, modeled the function as
y = 422.110(0.48896)^x
For x=7, this model gives y = 2.82
Using the first and last data points only, the model is
y = 200(0.06)^((x-1)/4)
For x=7, this model gives y = 2.94
Using yet another calculator to do linear regression on the log of y, the function is modeled by
y = 10^(2.62543-0.31073x)
which is another way to write the function produced by the TI-84 work-alike.
So, the technology I used gives a couple of reasons to choose
A. 2.86
and a couple of reasons to choose
C. 2.95
Of the models shown here, the ones with the lowest mean absolute deviation and the least squared error were the ones that produced the highest values of y for x=7, suggesting 2.95 is the best choice.
It would be advisable to use the procedure described in your text.
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Answer:
See below.
Step-by-step explanation:
See the attached chart that shows all outcomes for two dice.
There are 36 possible outcomes, all equally likely (if the dice are fair!).
The probability of getting a sum of 2 is 1/36 (only one way to get 2: both dice show 1).
The probability of getting a sum of 3 is 2/36 = 1/18.
The probability of getting a sum of 7 is 6/36 = 1/6 (the most likely sum!)
(Notice how all the sum = 7 outcomes like on a diagonal in the chart?)
The probability of getting a sum of 11 is 2/36 = 1/18.
The probability of getting a sum of 12 is 1/36.
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