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Klio2033 [76]
2 years ago
8

The area of a rectangular field is 2275 m2. The field is enclosed by 200 m of fence. What are the dimensions of the field?

Mathematics
1 answer:
AURORKA [14]2 years ago
4 0
So it is enclosed by 200 m of  fence
perimiter=200 m
it is a rectangle to
p=legnth+legnth+width+width or
p=l+l+w+w or
p=2l+2w or
p=2(l+w)
so
p=200
200=2(l+w)
divide both sides by 2
100=l+w
area=l times w
area=2275
lw=2275
l+w=100
combine and solve
l+w=100
subtract w
l=100-w
subisutute 100-w for l in other eqution
(100-w)(w)=2275
distribute
-w^2+100w=2275
add (w^2-100w) to both sides
0=w^2-100w+2275
factor
find what 2 numbers add up to -100 and multiply to get 2275
guess (or factor 2275 and find factors that add up to -100)
figure out that they are -65 and -35
0=(w-65)(w-35)
set each to zero
0=w-65
0=w-35

solve for w
w=65 or 35
65>35 so
65 m=legnth
35 m=width
 
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Given

p \to close

q \to fail to close

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This means that he closes the first three attempts. The event is represented as: p p p q

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Solving (b): He closes for the first time on the 3rd attempt

This means that he fails to close the first two attempts. The event is represented as: q q p

So, we have:

Pr = q * q * p

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Solving (c): First he closes is his 2nd attempt

This means that he fails to close the first. The event is represented as: q p

So, we have:

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Pr' = 1- 0.40^3

Pr' = 1- 0.064

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2 years ago
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Answer:

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