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3 years ago

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The perimeter of a rectangle is 400 feet and the length of the rectangle is 3 times the width. find the dimensions of the rectan

gle

1 answer:

bazaltina [42]3 years ago

8 0

Width: x
Length: 3x
Perimeter=x+x+3x+3x=8x
8x=400
x=50
Dimension is 150×50 feet (length × width)

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Simplify 3n+6(-8n+9)

Sedbober [7]

Step-by-step explanation:

3n + 6( -8n + 9)

= 3n - 48n + 54

= - 45n + 54

Hope it will help

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8/15 divided by 1/3 reduce your answer

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What are the slope between (-5,-2) and (8,13)

Cerrena [4.2K]

Answer:

The slope:

$m = \frac{15}{13}$

Step-by-step explanation:

-To determine the slope of the following points shown, you need this formula:

$m = \frac{y_{2} - y{1}}{x_{2} - x_{1}}$

( $m$ represents the slope, $(x_{1}, y_{1})$ represents the first coordinate and $(x_{2},y_{2})$ represents the second coordinate)

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Then, solve:

$m = \frac{13 + 2}{8 + 5}$

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3 years ago

Suppose each of the following data sets is a simple random sample from some population. For each dataset, make a normal QQ plot.

adell [148]

Answer:

a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.

c) This last case the histogram is not symmetrical and the data seems to be skewed.

Step-by-step explanation:

For this case we have the following data:

(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

We can use the following R code to get the histogram

> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)

> hist(x1,main="Histogram a)")

The result is on the first figure attached.

For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution

(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)

> hist(x2,main="Histogram b)")

The result is on the first figure attached.

For this case the data is skewed to the left and we can't assume that we have the normality assumption.

(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)

> hist(x3,main="Histogram c)")

The result is on the first figure attached.

This last case the histogram is not symmetrical and the data seems to be skewed.

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3 years ago