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antiseptic1488 [7]
3 years ago
6

HELP!!!!!! Which point approximates V3? A) A B) B C C D D

Mathematics
1 answer:
Igoryamba3 years ago
6 0

Answer:

c

ssusjs8snwi3heene8je

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Pleas help!!!! Brainliest and points!!
andreev551 [17]

Answer:

Step-by-step explanation:

Hi!

Here is the formula for point slope form: y-y1=m(x-x1)

Our slope in this case is 5, and our points are (1, 9). 1(x1) and 9(y1)

So, if we substitute our given data into the formula, here is what we will get.

y-9=5(x-1)

Thus, our answer should be:

y-9=5(x-1)

I hope this helps!

:)

5 0
3 years ago
Read 2 more answers
Y-intercept. y = -3/2x
Vaselesa [24]

Answer:

i hope you find the answer but unfortunaatly im not sure

Step-by-step explanation: thankyou for the points though :)

3 0
3 years ago
Three more than twice the sum of a number and half the number equals 105. Which equation models this relationship?
AlekseyPX

Answer:2(x+1/2x)+3=105

Step-by-step explanation: that is how i got my answer. If not a similar answer i would go with 2x+3=3x+105.

You can automatically eliminate the first answer. The second one has a random 5, so eliminated. and the last one just isn't correct.

7 0
3 years ago
What a pear weigh 200 kilograms are 200 grams
iren [92.7K]
200,000 milligrams is the correct answer
8 0
3 years ago
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Can you help me with number 6? <br> Confused abit <br> Please
Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.


To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.


First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is


\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}


Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is


\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}


Third question:

To get exactly one six, it can either be the first, second or third roll.


In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.


In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus


\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}


And since there are three of these combinations, The answer is


3\frac{5^2}{6^3}


Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is


1 - \frac{5^3}{6^3}

4 0
3 years ago
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