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Morgarella [4.7K]
3 years ago
11

What are the coordinates of the midpoint of AB IF A (6,4) and B (-8,8)

Mathematics
1 answer:
marissa [1.9K]3 years ago
3 0

Answer:

-I haven't done this, but I have copied and pasted from another question and answer from the user syed514:

Graphing is one way to do the problem.But sometimes, graphing it is hard to do.So here’s an algebraic method.

If M(m1, m2) is the midpoint of two points A(x1, y1) and B(x2, y2),then m1 = (x1 + x2)/2 and m2 = (y1 + y2)/2.In other words, the x-coordinate of the midpointis the average of the x-coordinates of the two points,and the y-coordinate of the midpointis the average of the y-coordinates of the two points.

Let B have coordinates (x2, y2) in our problem.Then we have that 6 = (2 + x2)/2 and 8 = (3 + y2)/2.

Solving for the coordinates gives x2 = 10, y2 = 13

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Your teacher gives you a number cube with numbers 1-6 on its faces. You are asked to state a theoretical probability model for r
butalik [34]

The missing part of the question is show in bold format.

Your teacher gives you a number cube with numbers 1-6 on its faces. You are asked to state a theoretical probability model for rolling it

once. Your probability model shows a probable outcome of 1/6 for each of the numbers on the cube, 1 chance for all any of the 6 numbers.

You roll it 500 times and get the following data:

Outcome 1 2 3 4 5 6

Frequency 77 92 75 90 76 90

Exercises 1–2

1. If the equality model was correct, about how many of each outcome

would you expect to see if the cube is rolled 500 times

2. Based on the data from the 500 rolls, how often were odd numbers observed? How often were even numbers observed?

Answer:

Step-by-step explanation:

1.

If the equally likely model was correct,  about how many of each outcome

would you expect to see if the cube is rolled 500 times.

The probability of rolling any of the numbers from 1 to 6 is  p(1/6)

The number of each of the outcomes expected to be seen in 500 rolls of the number cube is np

= 500 * \frac{1}{6}  \\ \\ = \frac{500}{6} \\ \\  = 83.333 \\ \\ \approx 83

2.  From the given data in the roll;

The odd numbers 1, 3 and 5, were obtained at  77, 75 and 76 times respectively.

Thus, the total number of times odd number were rolled = 77 + 75 + 76 = 228

Probability of an odd number turning up = \frac{ number \ of  \ required  \ outcome}{ total \  number  \ of  \ possible \ outcome}

= \frac{228}{500}

= 0.456

= 45.6%

The even numbers, 2, 4 and 6, were obtained 92, 90 and 90 times respectively.

The total number of times even number were rolled = 92 + 90 + 90 = 272

Probability of an even number turning up  = \frac{ number \ of  \ required  \ outcome}{ total \  number  \ of  \ possible \ outcome}

= \frac{272}{500}

= 0.544

= 54.4%

4 0
3 years ago
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