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Ronch [10]
3 years ago
6

Can you please answer the question?

Mathematics
1 answer:
Keith_Richards [23]3 years ago
3 0

Answer:

I think it is (B)

Step-by-step explanation:

I hope that is useful for you

Tell what happens in the comment please :)

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Suppose that 600 yards of fencing material are available to fence in
postnew [5]

Answer:

Length = 150 yards

Width = 100 yards

Step-by-step explanation:

We want 600 yards of fencing that will result in the largest 2 fenced corrals, sharing a common border.

It will take the shape of a rectangle, with a dividing fence down the center.

Let W and L,  Width and Length of the larger enclosure.

See attachment.

W= Area of the larger enclosure.

The perimeter is 2W + 2L.

The dividing fence is 1W

We know that we only have 600 yards of fence, so:

2W + 2L + 1W = 600 yards

Area = W x L

---

3W + 2L  = 600    (yards)

2L  = 600 -3W

L = (600-3W)/2

L = 300 -(3/2)W

---

Use this expression in the Area calculation:

Area = W x L

Area = W x (300 -(3/2)W)

Area = 300W -(3/2)W^2)

To find the maximum area, take the first derivative and set to zero to find the value of W that results in the greatest area.

Area' = 300 -2(3/2)W)

0 = 300 - 3W

3W = 300

W = 100 yards

Since 3W + 2L  = 600

L = (600 - 3W)/2

L = (600 - 3(100))/2

L = 150 yards

Area = 150*100 = 15,000 yards^2

 

8 0
3 years ago
hello loves!! I've been stuck on this question for like 2 hours lol! i need some help, ill give 100 points :)
lana66690 [7]

Answer:

61.12 units²

Step-by-step explanation:

(½ × pi × r²) + (½ × b × h)

½ [(3.14 × 4²) + (8 ×9)]

½(122.24)

61.12 units²

6 0
4 years ago
Read 2 more answers
Let $\mu$ and $\sigma^2$ denote the mean and variance of the random variable x. determine $e[(x-\mu)/\sigma]$ and $e{[((x-\mu)/\
Nezavi [6.7K]
\mathbb E\left(\dfrac{X-\mu}{\sigma}\right)=\dfrac1\sigma\mathbb E(X)-\dfrac\mu\sigma=\dfrac{\mu-\mu}\sigma=0

\mathbb E\bigg(\left(\dfrac{X-\mu}\sigma\right)^2\bigg)=\dfrac1{\sigma^2}\mathbb E\left(X^2-2\mu X+\mu^2\right)=\dfrac{\mathbb E(X^2)-2\mu\mathbb E(X)+\mu^2}{\sigma^2}
=\dfrac{\mathbb E(X^2)-2\mu^2+\mu^2}{\sigma^2}=\dfrac{\mathbb E(X^2)-\mu^2}{\sigma^2}=\dfrac{\mathbb E(X^2)-\mathbb E(X)^2}{\sigma^2}
=\dfrac{\mathbb V(X)}{\sigma^2}=\dfrac{\sigma^2}{\sigma^2}=1
3 0
4 years ago
HELP ASAP QUESTION IN PHOTO
cluponka [151]

Your answer would be A =7

8 0
3 years ago
What is this Help plz !
Arlecino [84]
24 square units is the answer
7 0
3 years ago
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