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Dahasolnce [82]

1 year ago

8

Rectangle KLMN is graphed on the coordinate plane below.The figure is rotated 360° counterclockwise using the origin as the cent

er of rotation. What is the location of the image of point K?
A. (2, –2)
B. (2, 2)
C. (–2, 2)
D. (–2, –2)

2 answers:

VladimirAG [237]1 year ago

3 0

Since 360 is a full loop around the circle, the points end up in the exact same place.

vazorg [7]1 year ago

3 0

Answer: The correct option is (A) (2, -2).

Step-by-step explanation: Given that rectangle with vertices at K, L, M and N is graphed on the coordinate plane as shown in the figure.The figure is rotated 360° counterclockwise using the origin as the center of rotation.

We are to find the location of the image of point K after the rotation.

We know that

a rotation of 360° in counterclockwise direction about the origin map a point to itself. That is, the co-ordinates of the image point will be same as the original one.

From the figure, we see that the co-ordinates of point K are (2, -2).

Therefore, after rotation of 360° counterclockwise using the origin as the center of rotation,

the co-ordinates of the image of point K will also be (2, -2).

Thus, the location of the image of point K is (2, -2).

Option (A) is CORRECT.

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Diana sold 336 muffins at the bake sale.Bob sold 287 muffins.Bob estimates that he sold 50 fewer muffins than Diana.how did he e

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Step-by-step explanation:

Since we have given that

Number of muffins Diana sold at the bake sale = 336

Number of muffins Bob sold at the bake sale = 287

Now, here we are using the method of estimation i.e. rounding the integers to nearest ones.

By estimation, we get

Number of muffins Diana sold at the bake sale = 340

Number of muffins Bob sold at the bake sale = 290

So,

Difference between them is given by

$340-290\\=50$

Hence, Bob sold 50 fewer muffins than Diana.

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Suppose that only 20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions whe

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Answer:

a) 91.33% probability that at most 6 will come to a complete stop

b) 10.91% probability that exactly 6 will come to a complete stop.

c) 19.58% probability that at least 6 will come to a complete stop

d) 4 of the next 20 drivers do you expect to come to a complete stop

Step-by-step explanation:

For each driver, there are only two possible outcomes. Either they will come to a complete stop, or they will not. The probability of a driver coming to a complete stop is independent of other drivers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of all drivers come to a complete stop at an intersection having flashing red lights in all directions when no other cars are visible.

This means that $p = 0.2$

20 drivers

This means that $n = 20$

a. at most 6 will come to a complete stop?

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.2)^{0}.(0.8)^{20} = 0.0115$

$P(X = 1) = C_{20,1}.(0.2)^{1}.(0.8)^{19} = 0.0576$

$P(X = 2) = C_{20,2}.(0.2)^{2}.(0.8)^{18} = 0.1369$

$P(X = 3) = C_{20,3}.(0.2)^{3}.(0.8)^{17} = 0.2054$

$P(X = 4) = C_{20,4}.(0.2)^{4}.(0.8)^{16} = 0.2182$

$P(X = 5) = C_{20,5}.(0.2)^{5}.(0.8)^{15} = 0.1746$

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

$P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 + 0.1091 = 0.9133$

91.33% probability that at most 6 will come to a complete stop

b. Exactly 6 will come to a complete stop?

$P(X = 6) = C_{20,6}.(0.2)^{6}.(0.8)^{14} = 0.1091$

10.91% probability that exactly 6 will come to a complete stop.

c. At least 6 will come to a complete stop?

Either less than 6 will come to a complete stop, or at least 6 will. The sum of the probabilities of these events is decimal 1. So

$P(X < 6) + P(X \geq 6) = 1$

We want $P(X \geq 6)$ . So

$P(X \geq 6) = 1 - P(X < 6)$

In which

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0115 + 0.0576 + 0.1369 + 0.2054 + 0.2182 + 0.1746 = 0.8042$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.8042 = 0.1958$

19.58% probability that at least 6 will come to a complete stop

d. How many of the next 20 drivers do you expect to come to a complete stop?

The expected value of the binomial distribution is

$E(X) = np = 20*0.2 = 4$

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