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dimulka [17.4K]
3 years ago
9

A rectangular dog pen is constructed using a barn wall as one side and 60 meters of fencing for the other three sides. What is t

he maximum area of the dog pen?
Mathematics
1 answer:
Sever21 [200]3 years ago
3 0

Answer:

450 m^2

Step-by-step explanation:

let the length be y m (the single side)

let the width be x m

so according to the question

2x + y = 60

or, y = 60 - 2x

now, Area = xy

= x(60-2x)

= -2x^2 + 60x

d(Area)/dx = -4x + 60

= 0 for a max of area

-4x + 60 = 0

x = 15

then y = 30

the width has to be 15 m, and the length has to be 30 m for a maximum area of 15(30) or 450 m^2.

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A rectangle has sides that measure x + 1/4
maksim [4K]

Answer:

6x + 5/6

Step-by-step explanation:

2(x + 1/4) + 2(2x + 1/3)

2x + 1/2 + 4x + 2/3

6x + 5/6

6 0
2 years ago
John wants to know the volume of his gold ring in cubic centimeters. He gets a glass in the shape of a rectangular prism with a
const2013 [10]
If you go to sc and slide to filter you’ll find one that solves the problem
7 0
3 years ago
I REALLY NEED HELP TO PASS!
expeople1 [14]
A = 1

4/17=2/8.5
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7 0
2 years ago
Read 2 more answers
Can you please show me the steps and solution to this problem?
ANEK [815]
M = - 7 - 3 / - 2 - 4
m = - 10 / - 6
m = 10/6
m = 5/3

y = 5/3x + c , (4,3)
3 = 5/3(4) + c
12 = 20 + 4c
4c = - 8
c = - 2
y = 5/3x - 2

i am a mathematics teacher. if anything to ask please pm me
4 0
2 years ago
Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.5 and a mean diameter of 205
Crank

Answer:

P(205-0.3=204.7

z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778

z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778

So we can find this probability:

P(-1.778

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

P = 1-0.9243 = 0.0757

Step-by-step explanation:

Let X the random variable that represent the diamters of interest for this case, and for this case we know the following info

Where \mu=205 and \sigma=1.5

We can begin finding this probability this probability

P(205-0.3=204.7

For this case they select a sample of n=79>30, so then we have enough evidence to use the central limit theorem and the distirbution for the sample mean can be approximated with:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}

And we can find the z scores for each limit and we got:

z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778

z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778

So we can find this probability:

P(-1.778

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

P = 1-0.9243 = 0.0757

6 0
2 years ago
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