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3 years ago

8

You have 80 yards of fencing to enclose a rectangular region find the dimensions of the rectangle that maximize the enclosed are

a

1 answer:

harkovskaia [24]3 years ago

3 0

A maximum area is always given by a square

so each side will be 80/4 = 20 yards long

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A group of three undergraduate and five graduate students are available to fill certain student government posts. If four studen

creativ13 [48]

Answer:

$Pr = 0.4286$

Step-by-step explanation:

Given

Let

$U \to\\$ Undergraduates

$G \to$ Graduates

So, we have:

$U = 3; G =5$ -- Total students

$r = 4$ --- students to select

Required

$P(U =2)$

From the question, we understand that 2 undergraduates are to be selected; This means that 2 graduates are to be selected.

First, we calculate the total possible selection (using combination)

$^nC_r = \frac{n!}{(n-r)!r!}$

So, we have:

$Total = ^{U + G}C_r$

$Total = ^{3 + 5}C_4$

$Total = ^8C_4$

$Total = \frac{8!}{(8-4)!4!}$

$Total = \frac{8!}{4!4!}$

Using a calculator, we have:

$Total = 70$

The number of ways of selecting 2 from 3 undergraduates is:

$U = ^3C_2$

$U = \frac{3!}{(3-2)!2!}$

$U = \frac{3!}{1!2!}$

$U = 3$

The number of ways of selecting 2 from 5 graduates is:

$G = ^5C_2$

$G = \frac{5!}{(5-2)!2!}$

$G = \frac{5!}{3!2!}$

$G =10$

So, the probability is:

$Pr = \frac{G * U}{Total}$

$Pr = \frac{10*3}{70}$

$Pr = \frac{30}{70}$

$Pr = 0.4286$

5 0

3 years ago

What does the perpendicular bisector of a line mean?

ankoles [38]

This bisector created a 90 degree angle

7 0

3 years ago

Read 2 more answers

A family went out to dinner and the cost of the food was $80. If the tip given was 15%, what

barxatty [35]

It should be $92.00

80 • .15 = 12
so 15% of the total ($80) is 12.

add the 12 to the total 80+12= $92

8 0

3 years ago

Read 2 more answers

suppose you invest 400 at an annual interest rate of 7.6% compounded continuously. How much will you have in the account after 1

dlinn [17]

A=pe^rt
A=400×e^(0.076×1.5)
A=448.30

3 0

3 years ago

Read 2 more answers

In tossing four fair dice, what is the probability of tossing, at most, one 3

11Alexandr11 [23.1K]

<u>Answer- </u>

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

<u>Solution-</u>

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

$P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48$ ( ∵ xxxx )

$P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38$ ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86

7 0

3 years ago