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2 years ago

A rotation R takes A(1,-3) to A'(3,5) and B(0,0) to B'(4,-6). find the centre of rotation?

1 answer:

6 0

The center of rotation will be (0,4). The points are plotted and the center is found with the help of a graph.

<h3>What exactly is a circle?</h3>

It is a point locus drawn equidistant from the center. The radius of the circle is the distance from the center to the circumference.

Given points;

A(1,-3)

A'(3,5)

B(0,0)

B'(4,-6)

C is the center point of rotation.

The given points are plotted on the graph and obtained as the center of rotation will be (0,4).

Hence the center of rotation will be (0,4).

To learn more about the circle, refer to the link: brainly.com/question/11833983.

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The computer that controls a bank's automatic teller machine crashes a mean of 0.6 times per day. What is the probability that,

professor190 [17]

Answer:

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Mean of 0.6 times a day

7 day week, so $\mu = 7*0.6 = 4.2$

What is the probability that, in any seven-day week, the computer will crash less than 3 times? Round your answer to four decimal places.

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4.2}*(4.2)^{0}}{(0)!} = 0.0150$

$P(X = 1) = \frac{e^{-4.2}*(4.2)^{1}}{(1)!} = 0.0630$

$P(X = 2) = \frac{e^{-4.2}*(4.2)^{2}}{(2)!} = 0.1323$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0150 + 0.0630 + 0.1323 = 0.2103$

0.2103 = 21.03% probability that, in any seven-day week, the computer will crash less than 3 times.

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3 years ago

Simplify i31. <br><br> 1 <br> −1<br> i<br> −i

Ede4ka [16]

The answer would be -i

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4 years ago

Part C of Jim and his dad are building a rectangular flower bed.

charle [14.2K]

I gotchu

The perimeter is 35. If we were to change the width, which is one of the dimensions of the flower bed, The perimeter will change. This means that perimeter will no longer be 35. So in order to keep the perimeter as it is, if we change one dimension, we must also change the other.

Let's solve for the length, using the formula to see how much the length changes from.

p = 2l + 2w

35 = 2l + 2(15)

35 = 2l + 30

5 = 2l

2.5 = l

We must increase the length from 2.5 feet. This is because decreasing one dimension will decrease the perimeter. But if we increase the other dimension as well, it will restore the perimeter to where is was initially.

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3 years ago

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What term must we add to complete the square on x^2 + 12x

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Answer:

Step-by-step explanation: