Lines k and p are perpendicular, neither is vertical and p passes through the origin. Which is greater? A. The product of the sl
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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.
Answer:
A junk box in your room contains fourteen old batteries, seven of which are totally dead.
So, number of good batteries =
a) The first two you choose are both good.
There is a 7/14 chance that we will pick a good battery.
Now there are 13 batteries left and out of that there are only 6 good ones remaining, so this becomes 6/13.
So, combined probability is =
= 0.23
b) At least one of the first three works.
And at least one is good battery, we get :
c) The first four you pick all work.
There is a probability of 7/14 for the first one to work, 6/13 for the second, 5/12 for the third and 4/11 when the fourth is good, combined we get by multiplying all:
d) You have to pick five batteries to find one that works.
This condition means that we pick 4 bad batteries and 1 good battery.
The probability of picking 4 bad batteries is -
Since there are 7 good batteries remaining in 10 batteries so we will multiply 7/10 in 0.0344 to know the fifth one that finally works.
This becomes =