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V125BC [204]
2 years ago
13

Find the slope and y intercept.​

Mathematics
2 answers:
FromTheMoon [43]2 years ago
5 0

Answer:

the slope is 2

the y intercept is (-4,0)

Step-by-step explanation:

slope is calcuated by rise over run

juin [17]2 years ago
4 0
The slope is 2
Y intercept is (-4,0)
Rise over run
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3 years ago
Please show me step by step how to solve this problem 3(-8+4v)=8v ?
Nadusha1986 [10]
So we know that:
3(-8 + 4v) = 8v
To find v, first simplify the left hand side:
-24 + 12v = 8v
Then group the "v's" over to the right:
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So v = 6

Hope this helped
8 0
3 years ago
There are 8 students on the minibus.Five of the students are boys. What fraction of the students are boys
Nimfa-mama [501]
Since 5 of the 8 students are boys, the fraction would be 5/8. The fraction would roughly be around 63%
5 0
3 years ago
Read 2 more answers
You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of
Dima020 [189]

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}

P_{1} is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is \frac{30}{50} = \frac{3}{5}.

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is \frac{33}{53}

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is \frac{36}{56}

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is \frac{39}{59}. So

P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588

P_{2} is the probability that we go R - R - B - R

P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788

P_{3} is the probability that we go R - B - R - R

P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076

P_{4} is the probability that we go R - B - B - R

P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511

P_{5} is the probability that we go B - R - R - R

P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733

P_{6} is the probability that we go B - R - B - R

P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493

P_{7} is the probability that we go B - B - R - R

P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476

P_{8} is the probability that we go B - B - B - R

P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419

So, the probability that this last marble is red is:

P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

P = P_{1} + P_{2}

P_{1} is the probability that we go R-R-R-R. It is the same P_{1} from the previous item(the last marble being red). So P_{1} = 0.1588

P_{2} is the probability that we go B-B-B-B. It is almost the same as P_{8} in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490

P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078

There is a 20.78% probability that we actually drew the same marble all four times

3 0
3 years ago
What is the answer? please help
amid [387]

Answer:

<h2><em>(</em><em>4</em><em>,</em><em>5</em><em>)</em></h2>

<em>sol</em><em>ution</em><em>,</em>

<em>A</em><em>(</em><em>2</em><em>,</em><em>7</em><em>)</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>></em><em> </em><em>(</em><em>X1,</em><em>y1</em><em>)</em>

<em>B</em><em>(</em><em>6</em><em>,</em><em>3</em><em>)</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>-</em><em>></em><em>(</em><em>x2</em><em>,</em><em>y2</em><em>)</em>

<em>now</em><em>,</em>

<em>ab = ( \frac{x1 + x1}{2}  \: , \frac{y1 + y2}{2} ) \\  =  (\frac{2 + 6}{2}  \: , \frac{7 + 3}{2} ) \\  = ( \frac{8}{2}  \:,  \frac{10}{2} ) \\  = (4 ,\: 5)</em>

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>

<em>Good</em><em> </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>

7 0
3 years ago
Read 2 more answers
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