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3 years ago

A student answered 44 questions correctly on a test with 55 questions. what percent of the test was answered correctly?

Mathematics

2 answers:

baherus [9]3 years ago

8 0

Answer:

Step-by-step explanation:

44 out of 55

divide :44/55=0.8

to convert it to decimal you multiply by 100 and put percentage next to it

0.8*100=80

%80

Zina [86]3 years ago

4 0

Answer:

80%

Step-by-step explanation:

The basic idea of this problem is to convert $\frac{44}{55}$ into a percentage. $\frac{44}{55}$ can be simplified to $\frac{4}{5}$ which is also 80%.

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Solve the inequality. −12−4x≤

disa [49]

Answer:

-4(3+x)

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find the value of two numbers is their sum is 21 and their difference is 3

kobusy [5.1K]

Hello from MrBillDoesMath

Answer: 12, 9

Discussion:

Let x be the first number and y the second number. The form the problem statement:

x + y = 21

x - y = 3

Add the two equations. This gives:

(x + x ) + ( y-y) = 21 + 3 =>

2x + 0 = 24 =>

x = 12

Since x -y =3, substituting in the value of x gives, 12 - y = 3 so y = 9

Thank you,

Mr. B

8 0

2 years ago

Read 2 more answers

A. 313 B. 133 C. 47 D. 227

stepan [7]

360 - 47 = 313

The answer is A

8 0

2 years ago

A sphere of radius r is cut by a plane h units above the equator, where

Anika [276]

Consider the top half of a sphere centered at the origin with radius $r$ , which can be described by the equation

$z=\sqrt{r^2-x^2-y^2}$

and consider a plane

$z=h$

with $0$ . Call the region between the two surfaces $R$ . The volume of $R$ is given by the triple integral

$\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates will help make this computation easier. Set

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

Now, the volume can be computed with the integral

$\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta$

You should get

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$

5 0

3 years ago

2 divided by 9 as a fraction help

Vera_Pavlovna [14]

2/9 is your answer
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4 0

3 years ago