3 years ago

use the formula P=2l+2w to find the length l of a rectangular lot if the width w is 45 feet and the perimeter P is 240 feet

Mathematics

1 answer:

Firdavs [7]3 years ago

5 0

240=2l+2(45) solve for l
240=2l+90 240-90=150/2=75

You might be interested in

What are TWO different ways you know when you are finished dividing?

vfiekz [6]

You know you are finished dividing when you can no longer simply & when you plug it back in to check the answer

7 0

2 years ago

Sebastian and two friends pooled their money and bought eight packages of baseball cards . there were six cards in each package.

zlopas [31]

8 • 6 = 48 cards total
48 ÷ 3 = 16 cards per person

7 0

2 years ago

A cylindrical container has a volume of 36 cm³. Another container made out of the same material is similar to the first containe

USPshnik [31]

I hope this helps you

Volume = pi.r^2.h

36= pi.r^2.h

Volume =pi.r^2.3/4h

Volume =pi.r^2.h.3/4

Volume =36.3/4

Volume =9.3

Volume =27

3 0

2 years ago

Read 2 more answers

3x+7=7x+8-3x how do i solve this? is it like infinite solutions, no solutions (Ø) or an actual number answer?

oksano4ka [1.4K]

Answer:

x=-1

Step-by-step explanation:

3x+7=7x+8-3x

3x+7=4x+8

-4x -4x

-x+7=8

-7 -7

-1

x=-1

8 0

1 year ago

Read 2 more answers

Assume we need to estimate the mean of a normally-distributed population with great accuracy. Specifically, for significance lev

bulgar [2K]

Answer:

$n = 2662.56\sigma^2$

Step-by-step explanation:

We are given the following in the question:

Significance level = 0.01

Width of interval = 0.1

Population variance = $\sigma^2$

We have to find the sample size so that the width of the confidence interval is no larger than 0.1

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.58$

Formula for sample size:

$n = \displaystyle\frac{z^2\sigma^2}{E^2}$

where E is the margin of error. Since the confidence interval width is 0.1,

$E = 0.05$

Putting these values in the equation:

$n = \displaystyle\frac{(2.58)^2\sigma^2}{(0.05)^2} = 2662.56\sigma^2$

So, the above expression helps us to calculate the sample size so that the width of the confidence interval is no larger than 0.1 for different sample variances.

4 0

3 years ago