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QveST [7]
2 years ago
6

The 225 students who passed all of their classes on the last report card will attend a one-hour Friday Fun-Day sponsored by the

Parent-Teacher
Organization (PTO). A variety of activities and snacks will be provided and students will receive one prize. A random survey of 50 students was
conducted to determine how many of each prize should be purchased for the Friday Fun-Day. The results of the survey are shown.
• 28 students want a slime kit.
• 12 students want a basketball.
• 6 students want a jewelry bead kit.
• 4 students want an artist sketch pad and pencil set.
Problem
Based on the survey results, which of these is NOT a valid prediction of the prizes wanted by the 225 students?
A There are 126 students who want a slime kit.
B There are 36 fewer students that want an artist sketch pad and pencil set than a basketball.
C There are 71 students who want either a basketball or a jewelry bead kit.
D There are 27 more students who want a slime kit than all the other prizes combined.
Mathematics
1 answer:
zhenek [66]2 years ago
8 0

Answer:

C. is incorrect.

Step-by-step explanation:

Calculate the percentages for each prize desired by the 50 students responding in the sample survey.  See the attached image for this calculation.  Multiply these percentages times the total student body of 225 students to determine the likely selection for each prize (on the attachment).

A.  is correct (126 students)

B.  is correct

C.  is incorrect.  The total is (54 + 27) = 83

D.  is true

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Sergeu [11.5K]
Total - book $ - magazine $

Which is equal to:

Total + (-book $) + (-magazine $)
$35 + (-$7) + (-$5)


ANSWER: (C) $35 + (-$7) + (-$5)

Hope this helps! :)
4 0
3 years ago
Help me pls pls help
Lelu [443]

Answer:

third one is a required answer.

x =2y

or

1/x=y

8 0
2 years ago
Read 2 more answers
Solve the equation.
mylen [45]

-9/15y+3/21=5/15y-14/21

Move 5/15y to the other side. Sign changes from +5/15y to -5/15y

-9/15y-5/15y+3/21=5/15y-5/15y-14/21

-14/15y+3/21=-14/21

Move 3/21 to the other side. Sign changes from +3/21 to -3/21.

-14/15y+3/21-3/21=-14/21-3/21

-14/15y=-14/21-3/21

-14/21-3/21=-17/21

-14/15y=-17/21

Multiply both sides by -15/14

-14/15y(-15/14)

Cross out 15 and 15, divide by 15 then becomes 1

Cross out 14 and 14, divide by 14 then becomes 1

1*1*y=y

-17/21*-15/14

Cross out 15 and 21 , divide by 3. 15/3=2, 21/3=7

17/7*5/14=85/98

Answer: c. y=85/98

7 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
I need help can you plz help me
horrorfan [7]

Simplify

10/3

Convert to a mixed fraction

3 1/3

4 0
3 years ago
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