OK. You asked for it. Here we go.
First, let's gather up the tools we might use ...
some things that we know about triangles:
-- Every triangle: Area = (1/2) x (length of the base) x (the height)
-- Isosceles triangle: It has two sides that are the same length.
-- Right triangle: It has one right angle in it.
The sides that meet at the right angle are called the "legs".
They form a corner there, like this _| .
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Now we can start using these tools to hack away at the problem.
Farmer Ted has an isosceles right triangle garden.
The problem asks us to figure out how long the legs are.
Before he changes anything, it looks like this _| and both of
those sides are the same length. Call it 'x' until we figure out
what it really is.
Notice that one of them is the base of the triangle, and the other one
is the height. So the area of this triangle is
(1/2) (x) (x) or (1/2) x² .
-------------------------------
Farmer Ted is never satisfied. Suddenly, one day without warning, he
comes along and makes the garden bigger. He makes one of the legs
7 ft longer, and he makes the other one 5 ft longer.
Now the length of one leg is (x + 7) and the other one is (x + 5) .
They're still the base and height of the triangle, so the area of the
bigger garden is
(1/2) (x + 7) (x + 5).
The problem says that this area is 55 square feet more than the original
area, so look out, here comes the <em>equation </em>:
new area = old area + bigger
(1/2) (x + 7) (x + 5) = (1/2) x² + 55
Locked in the mysterious shadowy crevices of this equation is
everything we need in order to figure out the original length of the legs ...
what we called 'x'.
At this point, we can forget about Farmer Ted, forget about the garden,
and just go back to our laboratory with this equation and solve it to find 'x'.
Let's take it slow and easy, one little step at a time:
<u>(1/2) (x + 7) (x + 5) = (1/2) x² + 55</u>
Multiply each side by 2 : (x + 7) (x + 5) = x² + 110
Expand (FOIL) the left side: x² + 12x + 35 = x² + 110
Subtract x² from each side: 12x + 35 = 110
Subtract 35 from each side: 12x = 75
Divide each side by 12 : <em>x = 6.25 feet</em>
==============================================
OK. That's a very nice number. How do we know whether it's correct ?
Let's check it out, and see if it fits the story:
Original area = (1/2 x base x height) = (0.5 x 6.25 x 6.25) = 19.53125 sq ft
One new leg = (6.25 + 7) = 13.25 ft
Other new leg = (6.25 + 5) = 11.25 ft
New area = (1/2 x base x height) = (0.5 x 13.25 x 11.25) = 74.53125 sq ft
How much bigger is the new area ?
74.53125 - 19.53125 = <em>55 sq ft </em> yay !
When we start with legs that are 6.25-ft and go through the whole story,
the new area is exactly what the problem says it was. So 6.25-ft is the
correct original length of the legs, before Farmer Ted messed with it.
Answer:
(A) 0.15625
(B) 0.1875
(C) Can't be computed
Step-by-step explanation:
We are given that the amount of time it takes for a student to complete a statistics quiz is uniformly distributed between 32 and 64 minutes.
Let X = Amount of time taken by student to complete a statistics quiz
So, X ~ U(32 , 64)
The PDF of uniform distribution is given by;
f(X) = 
, a < X < b where a = 32 and b = 64
The CDF of Uniform distribution is P(X <= x) = 
(A) Probability that student requires more than 59 minutes to complete the quiz = P(X > 59)
P(X > 59) = 1 - P(X <= 59) = 1 - = 1 - 
= 
= 0.15625
(B) Probability that student completes the quiz in a time between 37 and 43 minutes = P(37 <= X <= 43) = P(X <= 43) - P(X < 37)
P(X <= 43) = 
= 
= 0.34375
P(X < 37) = 
= 
= 0.15625
P(37 <= X <= 43) = 0.34375 - 0.15625 = 0.1875
(C) Probability that student complete the quiz in exactly 44.74 minutes
= P(X = 44.74)
The above probability can't be computed because this is a continuous distribution and it can't give point wise probability.
The domain is all the x values that makes the graph work so its -4,-1,3,5,6
