How many ways can 52 be arranged in?

jenna's rectangular garden borders a wall. she buys 80 ft of fencing. what are the dimensions of the garden that will maximize i

FrozenT [24]

Answer:

The dimensions are x =20 and y=20 of the garden that will maximize its area is 400

Step-by-step explanation:

Step 1:-

let 'x' be the length and the 'y' be the width of the rectangle

given Jenna's buys 80ft of fencing of rectangle so the perimeter of the rectangle is 2(x +y) = 80

x + y =40

y = 40 -x

now the area of the rectangle A = length X width

A = x y

substitute 'y' value in above A = x (40 - x)

A = 40 x - x^2 .....(1)

Step :2

now differentiating equation (1) with respective to 'x'

$\frac{dA}{dx} = 40 -2x$ ........(2)

Find the dimensions

$\frac{dA}{dx} = 0$

40 - 2x =0

40 = 2x

x = 20

and y = 40 - x = 40 -20 =20

The dimensions are x =20 and y=20

length = 20 and breadth = 20

Step 3:-

we have to find maximum area

Again differentiating equation (2) with respective to 'x' we get

$\frac{d^2A}{dx^2} = -2$

Now the maximum area A = x y at x =20 and y=20

A = 20 X 20 = 400

Conclusion:-

The dimensions are x =20 and y=20 of the garden that will maximize its area is 400

verification:-

The perimeter = 2(x +y) =80

2(20 +20) =80

2(40) =80

80 =80

8 0

3 years ago

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com

Rasek [7]

Answer:

a) The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

b)

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c) They are respected, as the upper bound of both intervals is below the new FAA recommendations.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve these questions.

Question a:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$ . So we have T = 1.9842

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9842\frac{20}{\sqrt{100}} = 4$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 183 - 4 = 179 pounds.

The upper end of the interval is the sample mean added to M. So it is 183 + 4 = 187 pounds.

The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

Question b:

Critical value is the same(same sample size and confidence level).

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.9842\frac{23}{\sqrt{100}} = 4.6$

The lower end of the interval is the sample mean subtracted by M. So it is 190 - 4.6 = 185.4 pounds.

The upper end of the interval is the sample mean added to M. So it is 190 + 4.6 = 194.6 pounds.

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

They are respected, as the upper bound of both intervals is below the new FAA recommendations.

7 0

3 years ago

Describe a real world problem that uses the expression 29 divided by (3/8 + 5/6).

Westkost [7]

29 divided by (3/8 + 5/6)

24

5 0

3 years ago

Qual o valor de Pi ?

Igoryamba

<h2>Hello my friend.</h2>

The Pi value is approximately equal to 3.14.