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Oksi-84 [34.3K]
3 years ago
15

This figure is dilated by a factor of 1 2 , with the origin as center. Which statement is NOT correct? A) R(2, 8) → R'(1, 4) B)

Q(10, 2) → Q'(5, 2) C) P(2, -4) → P'(1, -2) D) S(-10, 2) → S'(-5, 1)
Mathematics
2 answers:
Anton [14]3 years ago
7 0
Statement B would be the correct answer because the y's stayed the same. Having a dilation of 1/2 means that both the x and y from the original point will be multiplied by 1/2. Every other answer is multiplied by 1/2, meaning B is the correct answer. 

Hope this helps! If you need further assistance or explanation, message me or comment under this post!
k0ka [10]3 years ago
7 0

Answer:

Option B.

Step-by-step explanation:

If a figure is dilated by a factor of k , with the origin as center, then the rule of dilation is

(x,y)\rightarrow (kx,ky)

It is given that the given figure is dilated by a factor of 1 2 , with the origin as center. So, the rule of dilation is

(x,y)\rightarrow (\frac{1}{2}x,\frac{1}{2}y)

The vertices of image are

R(2,8)\rightarrow R'(1,4)

Q(10,2)\rightarrow Q'(5,1)

P(2,-4)\rightarrow P'(1,-2)

S(-10,2)\rightarrow S'(-5,1)

All the given statements are correct except Q(10, 2) → Q'(5, 2).

Therefore correct option is B.

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SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

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It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

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This probability is quite larger than 0.05.

Thus, the correct option is (A).

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