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Compare the reasoning as well as the answer. Equivalent valid methods are welcome. A referenced example provides a targeted repair route.
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Question 3
Guide question E3.3
The values 1, 4, and 7 occur with frequencies 2, 3, and 1, respectively. Find the mean of all six observations.
Enter an integer, decimal or fraction without units. Use up to 5 characters, or 6 including a leading minus.
Why this answer
3.5.
The total is 1(2) + 4(3) + 7(1) = 21, and the number of observations is 2 + 3 + 1 = 6. Thus the mean is 21/6 = 3.5. The unweighted mean of the three distinct values would be 4 and would answer a different question. A list method must repeat values according to their frequencies. Review Example 3.11.
An exact exponential model has outputs 80, 40, and 20 at times 0, 3, and 6. What does the model predict at time 9?
Enter an integer, decimal or fraction without units. Use up to 5 characters, or 6 including a leading minus.
Why this answer
10.
Every three time units, the output is multiplied by 1/2. The model is 80(1/2)t/3, so at t = 9 its value is 80(1/2)3= 10. The period is three units, not one. A regression template consistent with that period is y_1~a*b^(x_1/3). Review Example 3.08.
Write your complete response before opening the model answer. Compare both your answer and your reasoning.
Guide question E3.1
For f(x) = x2 − 2x + 3, make a two-row table for x = −1 and x = 2.
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f(−1) = 6andf(2) = 3.
Define f(x)=x^2-2*x+3. Enter −1 and 2 in the input column and use a computed output column f(x_1). The outputs are 1 + 2 + 3 = 6 and 4 − 4 + 3 = 3. Preserve row correspondence: swapping outputs would describe different ordered pairs. Review Example 3.01.
Guide question E3.2
Fit a least-squares line to the data (0, 1), (1, 4), (2, 4), and (3, 7). Use that fitted line to predict the output at x = 4.
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8.5.
With the pairs in columns x1,y1, use y_1~m*x_1+b. The fitted line is y = 1.8x + 1.3, which predicts 1.8(4) + 1.3 = 8.5. This is a fitted prediction, not an observed value. As a check, the line passes through the mean point (1.5, 4) because 1.8(1.5) + 1.3 = 4. Review Example 3.05.
Guide question E3.5
A quadratic y = ax2 + bx + c passes through (0, 2) and (1, 5). Is its value at x = 2 determined uniquely? Justify your answer.
Show model answerHide model answer for question E3.5
No.
The first point fixes c = 2, while the second gives a + b = 3. Both x2 + 2x + 2 and 2x2 + x + 2 satisfy the two points, but their outputs at 2 are 10 and 12. The data do not identify a unique three-parameter quadratic. A regression display cannot create the missing condition. Review Example 3.14.