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3 years ago

NEED DONE ASAP. IT'S 12AM AND I WANNA GET THIS DONE SO I CAN SLEEP..

Mathematics

1 answer:

mr Goodwill [35]3 years ago

8 0

The options seem a bit mis-formatted but here is the way p is computed:

the decrease is 2000-1600

and it is relative to: 2000 (2003 attendance)

so $p=100 \frac{2000-1600}{2000} = 100 \frac{400}{2000}$

and so that aligns with your Option A. Options B, C, D look incorrect for sure.

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Samara has an assignment to conduct a survey. She wants to find out what students like to do after school. She asks the first 10

Sliva [168]

C: Because she doesn't choose certain people

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4 years ago

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X=-12-6y <br>-4x+5y=-39<br>slove by the system of substitution

Artist 52 [7]

X = -6y - 12
4x + 5y =-39

4x + 5y = -39
-4(-6y - 12) + 5y = -39
-4(-6y) + 4(12) + 5y = -39
24y + 48 + 5y = -39
29y + 48 = -39
29y = -87
y = -3

x = -6y - 12
x = -6(-3) - 12
x = 28 - 12
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(x, y) = (6, -3)

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4 years ago

What is the area of the circle if DF=20 in.

crimeas [40]

We know that

DF is the diameter of the circle
so
radius r=DF/2----->20/2-----> r=10 in

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the answer is

100*pi in² or 314 in²

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3 years ago

Help please this is confusing and also show your work

liq [111]

Answer:

Step-by-step explanation:

Volume is length · width · height, so you can add another column with the volumes:

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3 years ago

Which point is in the solution set of the given system of inequalities 3x+y>-3,x+2y<4

Dima020 [189]

Step 1. Solve both inequalities for $y$ :
$3x+y\ \textgreater \ -3$
$y\ \textgreater \ -3x-3$

$x+2y\ \textless \ 4$
$2y\ \textless \ -x+4$
$y\ \textless \ - \frac{1}{2} x+2$

Step 2. To check a point in the solution of the given system of inequalities, look for the intercepts of the lines $-3x-3$ and $- \frac{1}{2} x+2$ :

$y=-3x-3$ (1)
$y=- \frac{1}{2} x+2$ (2)

Replace (1) in (2):
$-3x-3=- \frac{1}{2} x+2$
Solve for $x$ :
$\frac{5}{2} x=-5$
$x=-2$ (3)

Replace (3) in (1):
$y=-3x-3$
$y=-3(-2)-3$
$y=6-3$
$y=3$

We can conclude that the point (-2,3) is in the solution of the system if <span>inequalities</span>; also any point inside the dark shaded area of the graph of the system of inequalities is also a solution of the system.

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3 years ago

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