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storchak [24]
3 years ago
8

Point Farmer Ed has 60 feet of fencing, and wants to enclose a rectangular plot that borders on a river. If Farmer Ed does not f

ence the side along the river, find the length and width of the plot that will maximize the area. What is the largest area that can be enclosed?​
Mathematics
1 answer:
miskamm [114]3 years ago
7 0

Answer:

Farmer Ed has 60 feet of fencing; and wants to enclose rectangular plot that borders on river: If Farmer Ed does not fence the side along the river; find the length and width of the plot that will maximize the area_ What is the largest area that can be enclosed? What width will maximize the area? The width, labeled x in the figure. (Type an integer or decimal ) What length will maximize the area? The length, labeled in the figure, is (Type an integer or decimal ) What is the largest area that can be enclosed? The largest area that can be enclosed is (Type an integer or decimal.)

You have 120 feet of fencing to enclose a rectangular plot that borders on a river.

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O. Shannon's hair is 12 inches long and grows 0.25 inch per
wariber [46]

Answer:

0.25 (inches?)

Step-by-step explanation:

the equation is:

y=mx+b

y is the length of her hair, the dependent variable, the outcome

m is the slope, which is in this case the growth of her hair in a month

x is the independent variable, or the number of months

b is how long her hair is to start with, which is 12 in

we are looking for m, which is the growth of her hair per month, which is given as 0.25 inches, so that would be the answer

3 0
4 years ago
A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data
Rasek [7]

Answer: a. 1.981 < μ < 2.18

              b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

\overline{x}=\frac{\Sigma x}{n}

\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}

\overline{x}= 2.08

Now, we estimate standard deviation:

s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }

s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }

s = 0.1564

For t-score, we need to determine degree of freedom and \frac{\alpha}{2}:

df = 12 - 1

df = 11

\alpha = 1 - 0.95

α = 0.05

\frac{\alpha}{2}= 0.025

Then, t-score is

t_{11,0.025} = 2.201

The interval will be

\overline{x} ± t.\frac{s}{\sqrt{n} }

2.08 ± 2.201\frac{0.1564}{\sqrt{12} }

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

8 0
3 years ago
3 soccer balls and 5 more how many are there​
vodomira [7]

Answer:

8

Step-by-step explanation:

3+5=8

6 0
3 years ago
What is the solution to this system of linear equations?<br><br> x + y = 4<br><br> x − y = 6
Amiraneli [1.4K]
X+y = 4, therefore y = 4-x when you rearrange the equation.
You can subsitute this in: x-(4-x) = 6
-4+x = 6-x
x=10-x
2x = 10
x = 5
Substitute x into the previous equation:
5+y = 4
y = 4-5
y = -1
5 0
4 years ago
Read 2 more answers
What is the constant of variation, k, of the line y = kx through (3,18) and (5,30)?
Fudgin [204]
Y is 18 so 18=3k
k=18/3
k=6 for (3,18)
for (5,30)let's see
y=30
30=5k
k=30/5
k=6
5 0
3 years ago
Read 2 more answers
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