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schepotkina [342]
2 years ago
14

Find the slope of the line passing through the point (-6,8) and (-9,1)

Mathematics
1 answer:
Ipatiy [6.2K]2 years ago
5 0
Hello!

Y2-y1/x2-x1

1-9/8-6
-8/2

The slope is -4

Hope this helps!

-bambi
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(4p-4)+(-7p-7)<br> I need HELP ON THIS PLEASE FAST!!
yuradex [85]

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The correct answer is number 2

-3p - 11

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3 years ago
Explain how you can rename 5,400 as hundreds.
balandron [24]

Answer:

The correct answer would be 54 hundreds

Step-by-step explanation:

If we are considering how many hundreds something has, we can simply take away two 0's and use that number.  We take the two 0's off and then we have the right number of hundreds

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3 years ago
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The Bobcats had a win-loss record of 41-23 before their star player, Melissa, got injured. Melissa didn’t play in any more games
MrRissso [65]

Answer:

The win percentage decreased by 10%

Step-by-step explanation:

First you need to fin the total games played right before Melissa got Injured

41+23=64

Then you need to find the win percentage by making a fraction of wins over total games

wins/total games = 41/64 = 64%

Then you need to find the number of games played after Melissa got injured

54-41 = 13 wins

34-23 = 11 losses

13+11 = 24 total games played without Melissa

You find that win percentage similarly to the first time

wins/total games = 13/24 = 54%

Then you find the difference in the percentages and there you have it!!

64% - 54% = 10%

It decreased as well therefore the answer is...

It decreased by 10%

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3 years ago
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Step-by-step explanation:

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Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th
grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

S = \{1,1,2,2,2,3\}

And the probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{6}

P(1) = \frac{1}{3}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{6}

P(2) = \frac{1}{2}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{6}

P(Odd Number) is then calculated as:

P(Odd\ Number) =  P(1) + P(3)

P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}

Take LCM

P(Odd\ Number) = \frac{2+1}{6}

P(Odd\ Number) = \frac{3}{6}

P(Odd\ Number) =  \frac{1}{2}

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

S = \{1,1,2,2,2,3,3\}

The probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{7}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{7}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{7}

P(Odd\ Number) = P(1) + P(3)

P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}

Take LCM

P(Odd\ Number) = \frac{2+1}{7}

P(Odd\ Number) = \frac{3}{7}

Comparing P(Odd Number) before and after

P(Odd\ Number) =  \frac{1}{2} --- Before

P(Odd\ Number) = \frac{3}{7} --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0
3 years ago
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