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makvit [3.9K]
2 years ago
10

A farmer with 1670 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does

not fence the side along the highway, what is the largest area that can be enclosed? Give your answer correct to the nearest square meter.
Mathematics
1 answer:
anygoal [31]2 years ago
5 0

The largest area that can be enclosed to the nearest square meter will be 348613 m²

<u><em>Explanation</em></u>

Suppose, the length and width of the rectangular plot are  l and w respectively.

The farmer does not fence the side along the highway. Lets assume, the <u>farmer does not fence across one length side</u>. So, the total fence needed = l+2w

Given that, the length of the fence is 1670 meters. So, the equation will be.....

l+2w=1670\\ \\ l= 1670-2w

Now, the area of the plot.....

A= l*w\\ \\ A= (1670-2w)w \\ \\ A=1670w-2w^2

Taking derivative on both sides of the above equation in respect of w , we will get.....

\frac{dA}{dw}=1670-2(2w)\\ \\ \frac{dA}{dw}=1670-4w

<u>Now A will be maximum when \frac{dA}{dw}=0 </u>. So....

1670-4w=0\\ \\ 4w=1670\\ \\ w=\frac{1670}{4}=417.5

Thus, the area will be:  A= 1670w-2w^2=1670(417.5)-2(417.5)^2= 348612.5 \approx 348613

So, the largest area that can be enclosed to the nearest square meter will be 348613 m²

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miskamm [114]

Answer:

The probability is 28 % ( approx )

Step-by-step explanation:

Since, there are only two possible outcomes in every trails ( residents are in their forties or not ),

So, this is a binomial distribution,

Binomial distribution formula,

Probability of success in x trials,

P(x)=^nC_x p^x q^{n-x}

Where, ^nC_x=\frac{n!}{x!(n-x)!}

p and q are probability of success and failure respectively and n is the total number of trials.

Let X be the event of resident who are in his or her forties.

Here, p = 12% = 0.12,

⇒ q = 1 - p = 1 - 0.12 = 0.88,

n = 9,

Hence, the probability that two or three of the people in the group are in their forties = P(X=2) + P(X=3)

^9C_2 (0.12)^2 (0.88)^{9-2}+^9C_3 (0.12)^3 (0.88)^{9-3}

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=28\%

7 0
3 years ago
Both Need To Be Checked :-)
Tems11 [23]
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40 - 10 + 20
marin [14]

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Circle P is described by the equation (x−1)2+(y+6)2=9 and circle Q is described by the equation (x+4)2+(y+14)2=4. Select from th
xeze [42]

Answer:

(a) Circle Q is 9.4 units to the center of circle P

(b) Circle Q has a smaller radius

Step-by-step explanation:

Given

P:(x - 1)^2 + (y + 6)^2 = 9

Q:(x + 4)^2 + (y + 14)^2 = 4

Solving (a): The distance between both

The equation of a circle is:

(x - h)^2 + (y - k)^2 = r^2

Where

Center: (h,k)

Radius:r

P and Q can be rewritten as:

P:(x - 1)^2 + (y + 6)^2 = 3^2

Q:(x + 4)^2 + (y + 14)^2 = 2^2

So, for P:

Center = (1,-6)

r = 3

For Q:

Center = (-4,-14)

r = 2

The distance between them is:

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2

Where:

Center = (1,-6) --- (x_1,y_1)

Center = (-4,-14) --- (x_2,y_2)

So:

d = \sqrt{(1 - -4)^2 + (-6 - -14)^2

d = \sqrt{(5)^2 + (8)^2

d = \sqrt{25 + 64

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In (a), we have:

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2 years ago
What is the inverse of y=3x+3?
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Answer:

The inverse is x/3 -1

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y = 3x+3

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