In this situation, you have:
The mode is 79 (that’s the most common number).
The median is 78.5 (the middle value or the average of the two middle values, since there is an even number of test scores).
The mean is less than both of those, as it was brought down by the few scores of 49.
The means the mean will be less than median.
Answer: C
Answer:
Step-by-step explanation:
I think it is y=-.25
7*3 - 4y = 20
21 - 4y = 20
subtract 21 from each side
- 4y = - 1
divide each side by - 4
y= -.25
Answer:
-2, 8/3
Step-by-step explanation:
You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...
A = (1/2)(f(a) +f(4))(4 -a)
= (1/2)((3a -1) +(3·4 -1))(4 -a)
= (1/2)(3a +10)(4 -a)
We want this area to be 12, so we can substitute that value for A and solve for "a".
12 = (1/2)(3a +10)(4 -a)
24 = (3a +10)(4 -a) = -3a² +2a +40
3a² -2a -16 = 0 . . . . . . subtract the right side
(3a -8)(a +2) = 0 . . . . . factor
Values of "a" that make these factors zero are ...
a = 8/3, a = -2
The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.
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<em>Alternate solution</em>
The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.
It's subtraction or multiplacation