Your answer is going to be 55.76$
We are dealing here with a uniform distribution ranging from 0 to 30 minutes. We need to calculate the probability that the unreliable bus will arrive before the reliable one. This probability is the area under the uniform distribution "curve" from 0 to 10 minutes. This constitutes 1/3 of the entire unform distr. curve. So the probability that the unreliable bus will arrive before the reliable one is 1/3, or 0.33. The probability that it will arrive AFTER the reliable bus is 2/3, or 0.67.
Ummm... I explained the whole process earlier, and was mistakenly issued for plagiarism. Which really sucks, because it took forever to type and format all of those formulas and steps.
Here's the shortened version...
V=
d (diameter) = 2r
so 24/2 = 12
plug in...
You should get approx. 7234.56, and don't forget to label it appropriately, it should be units cubed.
Answer:
1. y = (1/6)(x +5)^2 -4.5
2. y = (-1/20)(x -10)^2 +1
Step-by-step explanation:
For focus (p, q) and directrix y=r, the equation is ...
y = 1/(2(q-r))·(x -p)^2 +(q+r)/2
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1. For (p, q) = (-5, -3) and r = -6, the equation is ...
y = 1/(2(-3-(-6))(x -(-5))^2 +(-3-6)/2
y = 1/6(x +5)^2 -4.5
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2. for (p, q) = (10, -4) and r = 6, the equation is ...
y = 1/(2(-4-6))(x -10)^2 +(-4+6)/2
y = (-1/20)(x -10)^2 +1