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elena-s [515]
3 years ago
6

Joe got a bonus of $1,600.20 based on a certain percent of his sales for the year, which was $22,860. What percent of his sales

for the year was the bonus?
Mathematics
1 answer:
tekilochka [14]3 years ago
3 0
If its rounded to nearest hundredth its 14.29%, and if rounded to nearest tenth its 14.3%
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The price of a car was increased from $3,600 to $4,500. What was the percent of the increase?
Nataly [62]

Answer:

25%

Step-by-step explanation:

.25(3,600)=900

3,600+900=4,500

6 0
3 years ago
Read 2 more answers
Pls answer this with a picture showing the correct box plot
faust18 [17]

Answer/Step-by-step explanation:

To find out the mistake of the student, let's find the min, max, median, Q1 and Q3, which make up the 5 important values that are represented in a box plot.

Given, {2, 3, 5, 6, 10, 14, 15},

Minimum value = 2

Median = middle data point = 6

Q1 = 3 (the middle value of the lower part of the data set before the median)

Q3 = 14 (middle value of the upper part of the data set after the median)

Maximum value = 15

If we examine the diagram the student created, you will observe that he plotted the median wrongly. The median, which is represented by the vertical line that divides the box, ought to be at 6 NOT 10.

See the attachment below for the correct box plot.

3 0
3 years ago
Which is a correct two-column proof?<br> Given: ∠h and ∠c are supplementary.<br> Prove: j || l
pentagon [3]

Answer:

Third option is the correct answer.

Step-by-step explanation:

Third option is the correct answer.

7 0
3 years ago
What is the perimeter of this red
Vilka [71]
96

multiply the side length by the number of sides
4 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
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