Use technology to construct an appropriate model of the data. (0,1), (1,-2), (2,-3), (3,-2), (4,1) f(x) = – x2 + 4 f(x) = x2 - 4
f(x) = - x2 + 4x - 1 f(x) = x2 - 4x + 1
a. f(x) = – (x+2)2 + 3B. f(x) = (x+2)2 + 3C. f(x) = – (x-2)2 - 3D. f(x) = (x-2)2 - 3
a. f(x) = – (x+2)2 + 3B. f(x) = (x+2)2 + 3C. f(x) = – (x-2)2 - 3D. f(x) = (x-2)2 - 3
1 answer:
The answer choices tell you this is a parabola.
The given points tell you it opens upward and the vertex is (2, -3).
Your knowledge of the vertex form of the equation is the only "technology" that you need. The vertex form is
y = a(x - h)² + k . . . . . for vertex (h, k)
The fact that the y-value is 1 more than its vertex value when |x-2| = 1 tells you the scale factor "a" is 1.
You can write the model as
D. f(x) = (x -2)² - 3
The given points tell you it opens upward and the vertex is (2, -3).
Your knowledge of the vertex form of the equation is the only "technology" that you need. The vertex form is
y = a(x - h)² + k . . . . . for vertex (h, k)
The fact that the y-value is 1 more than its vertex value when |x-2| = 1 tells you the scale factor "a" is 1.
You can write the model as
D. f(x) = (x -2)² - 3
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To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.
To calculate the degrees of freedom you need to use the following equation:
≈34
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.
To compare this two sets of data, you need to use a t-student test:
You have the following data:
-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph
-Wednesday n2=20; </span>x̄2=56,3 mph; s2=4,4 mph
You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.
To calculate the degrees of freedom you need to use the following equation:
The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10
So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.