All categories

3 years ago

A rectangular prism has a height of 20 cm, length of 7 cm and width of 5 cm.

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

7 0

Formula for volume is:

width x length x height

5 x 7 x 20

5x7 = 35
35x20= 700
Answer is 700 cm^2

You might be interested in

Ty uses 20 ft of fence to divide into 4 equal parts all in one row. Each aware part is 2 feet long. What is the area of the gard

stich3 [128]

Answer:

Step-by-step explanation:

20 divided by 4 = 5 and times it from there

3 0

2 years ago

The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f

WITCHER [35]

Answer:

Part 1) $A=36(\pi-2)\ cm^2$

Part 2) $P=6(\pi+2\sqrt{2})\ cm$

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

$A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)$

we have that the base and the height of triangle is equal to the radius of the circle

$r=12\ cm\\b=12\ cm\\h=12\ cm$

substitute

$A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2$

simplify

Factor 36

$A=36(\pi-2)\ cm^2$

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

$C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r$

substitute the given values

$C=\frac{1}{2}\pi (12)\\C=6\pi\ cm$

Applying the Pythagorean Theorem

The hypotenuse of right triangle is equal to

$AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm$

simplify

$AC=12\sqrt{2}\ cm$

Find the perimeter

$P=(6\pi+12\sqrt{2})\ cm$

simplify

Factor 6

$P=6(\pi+2\sqrt{2})\ cm$

4 0

3 years ago

In a sample of 42 water specimens taken from a construction site, 26 contained detectable levels of lead. Construct 90% confiden

hram777 [196]

Answer:

(0.4958, 0.7422)

Step-by-step explanation:

Let p be the true proportion of water specimens that contain detectable levels of lead. The point estimate for p is $\hat{p}=26/42=0.6190$ . The estimated standard deviation is given by $\sqrt{\hat{p}(1-\hat{p})/n}=\sqrt{(0.6190)(1-0.6190)/42}=0.0749$ . Because we have a large sample, the 90% confidence interval for p is given by $0.6190\pm z_{0.05}0.0749$ where $z_{0.05}=1.6448$ is the value that satisfies that above this and under the standard normal density there is an area of 0.05. So, the confidence interval is $0.6190\pm (1.6448)(0.0749)$ , i.e., (0.4958, 0.7422).