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eduard
3 years ago
7

What is question is asking

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
5 0
A) 72/910
b) 278/910
c) 57/910
d1) same as c) = 57/910
d2) 120/910
d3) 101/910
e) No, they DEPENDANT. 
57% of Liberals approve citizenship against 15% of conservatives and 33% of Moderates





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Tracy solved the following problem. Which of the statements is true?
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it would be the third answer choice, when doing this you always make sure that there is multiplication or division you do those first going left to right, even if there is addition first, you still would do either multiplication or division first.

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3 years ago
Lines A,B and C show proportional relationships which line has a constant of proportionally between y and x of 5
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34

Step-by-step explanation:

3/5/45

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Last year there were 20 girls on the volleyball team. This year the number of girls decreased by 10%. How many girls are on the
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Consider the given function.
Natali5045456 [20]

Answer:

D)

Step-by-step explanation:

f(x) = x^2 -14x -72

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asis of symmetry: x=(x1+x2)/2

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3 0
3 years ago
PLEASE HELP!!!
pogonyaev

Step-by-step explanation:

Part A:

Let m be the number of mittens and s be the number of scarves. Then we have the inequalities:

s+m\leq 30. <em>This says Nivyana and Ana cannot make more than 30 scarves</em>

50s+25m\geq 1000. <em>This says that</em> <em>Nivyana and Ana have to earn at least $1000.</em>

Part B:

The graph is attached.

Notice that the graphs of the inequalities are solid lines, this just means that the points on these lines included to the solutions of each inequality.

The darker shaded region and the solid lines bounding it, are the solutions to the inequalities because that's where the values common to both inequalities are found.

Part C:

From the graph we get two possible solutions:

15 scarves & 10 mittens

25 scarves & 5 mittens.

These two points lie on the solid lines that bound the darker shaded region<em> (I picked those points to stress that the lines bounding the dark region are also solutions.)</em>

8 0
3 years ago
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