The initial value is 3 (when x=0), and the multiplier is 2, when x=2. The equation can be written as
.. y = 3*(2^(x/2))
.. y = 3*(2^(x/2))
Graph the line with the equation y = -1/2x – 7
1 answer:
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Answer:
Five classes
Step-by-step explanation:
1. Sort the data
You get
1.1, 1.3, 1.6, 1.7, 2, 2.6, 2.7, 2.9, 3.2, 3.2, 3.5, 3.5, 3.9, 4.6, 4.7, 4.8, 4.8, 4.9, 4.9, 5.3, 5.7, 6.4, 6.5, 7.1, 7.5, 7.6, 8.1, 8.2, 9.2, 9.4
2. Calculate the range
Range = Max - Min = 9.4 - 1.1 = 8.3
3. Calculate the class width
Divide the range by the number of classes
8.3/5 = 1.7
Round this up to 2.
4. Decide where to start the histogram
You could use classes: 1 - 3, 3 - 5, 5 - 7, 7 - 9, 9 -11
However, even numbers are easier to read.
It would be preferable to use classes: 0 - 2, 2 - 4, 4 - 6, 6 - 8, 8 - 10
5. Prepare a frequency distribution table
Note: Each class does not include its largest possible value. Thus, a value of 2 goes into Class 2 - 4, (not Class 0 - 2).
Answer:
The correct options are;
Answer to A1 is D
Answer to A2 is D
Answer to A3 is D
Answer to A4 is D
Answer to A5 is D
Answer to A6 is D
Answer to A7 is D
Answer to A8 is D
Answer to A9 is D
Answer to B1 is I
Answer to B2 is I
Answer to B3 is I
Answer to B4 is I
Answer to B5 is I
Answer to B6 is I
Step-by-step explanation:
The given function is f(x) = 9·x² + 54·x - 66
The extremum of the function are found as follows;
d(f(x))/dx = 0 = d(9·x² + 54·x - 66)/dx = 18·x + 54
∴ 18·x + 54 = 0 at the maximum or minimum points
x = -54/18 = -3
Given that d²(f(x))/dx² = 18 > 0. x = -3 is a minimum point
Given that the function is a quadratic function, we have;
1) Points to the left of x = -3 are decreasing
2) Points to the right of x = -3 are increasing.
