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Anit [1.1K]
3 years ago
14

The length of a rectangular fenced enclosure is 12 feet more than the width. If Farmer Dan has 100 feet of fencing, write an ine

quality to find the dimensions of the rectangle with the largest perimeter that can be created using 100 feet of fencing.

Mathematics
2 answers:
Kitty [74]3 years ago
7 0
Answer:
length = 31 ft
width = 19 ft

Explanation:
Assume that the width of the rectangle is w.
We are given that the length is 12 ft more than the width. This means that the length of the rectangle is w + 12

The perimeter of the rectangle in this case would be:
perimeter = 2 (length + width)
perimeter = 2 (12 + w + w)
perimeter = 24 + 4w

Assume that Dan would use all 100 ft of fencing to surround the yard. This would mean that the largest perimeter is 100 ft.

Therefore:
perimeter = 24 + 4w
100 = 24 + 4w
4w = 100 - 24
4w = 76
w = 19 ft

Since we have calculated that the width of the yard is 19 ft, we can substitute to get the length as follows:
length = 12 + w
length = 12 + 19
length = 31 ft

Hope this helps :)
Lynna [10]3 years ago
4 0
To answer this question you will write an equation using 100 as the perimeter and w + 12 as the length.

See attached picture for this work.

The biggest width would be 19 feet. The inequality shows that the width would need to be at 19 feet or lower.

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Please answer all parts as I know the answers but need the work to go with them. Thus, I believe the above answers are correct.
vampirchik [111]

As the sample size n is less than 30 normal distribution is used.

The values of x are converted into z by using the formula z= x-u/ s and then the z values are found out from the table.

The limits are found by using the formula x±σz or x±sz where s= σ

As the sample size is 10 which is less than 30 the normal distribution is used.

The probability of x< 2.59 is 0.3446

The probability 2.60<X <2.63 is 0.9484

So lower and upper limits are 2.607 and 2.612

Part A

As the sample size is 10 which is less than 30 the normal distribution is used.

Part B

For given value of x= 2.59 z is obtained =0.4

x= 2.59

z= x-u/ s

z= 2.59-2.61/0.05

z= -0.02/0.05

z=- 0.4

P (X<2.59) = P(-0.4 <Z<0) = 0.5 -0.1554= 0.3446

The probability of x< 2.59 is 0.3446

Part C

For two given values of x= 2.60 and 2.63 z is obtained as =0.2 and 0.4

x1= 2.60

z1= x-u/ s

z= 2.60-2.61/0.05

z= -0.01/0.05

z=- 0.2

x2= 2.63

z2= x-u/ s

z= 2.63-2.61/0.05

z= 0.02/0.05

z= 0.4

P (2.60<X<2.63) = P(-0.2 <Z<0.4)

= P(-0.2 <Z<0)+ P(0 <Z<0.4)

=0.793 + 0.1554= 0.9484

The probability 2.60<X <2.63 is 0.9484

Part D"

p= 0.57

From the table z= 0.045

z= x-u/ s

zs= x-u

zs+u = x

x1= 0.045*0.05 +2.61= 2.61225

x2= 2.61- 0.00225= 2.60775

So lower and upper limits are 2.607 and 2.612

For further understanding of probability calculation click

brainly.com/question/25638875

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