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3 years ago

Determine whether each scale factor produces an expansion or a contraction under a dilation of the original image

Mathematics

1 answer:

blondinia [14]3 years ago

4 0

Answer:

see the explanation

Step-by-step explanation:

we know that

If the absolute value of the scale factor is less than 1, then the dilation produces a contraction of the original image

If the absolute value of the scale factor is greater than 1, then the dilation produces an expansion of the original image

<u><em>Verify each value</em></u>

1) -4

$\left|-4\right|=4$

$4> 1$

therefore

The dilation produces an expansion of the original image

2) 0.25

$\left|0.25\right|=0.25$

$0.25 < 1$

therefore

The dilation produces a contraction of the original image

3) -2/3

$\left|-2/3\right|=2/3$

$2/3 < 1$

therefore

The dilation produces a contraction of the original image

4) 2.3

$\left|2.3\right|=2.3$

$2.3 > 1$

therefore

The dilation produces an expansion of the original image

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3 years ago

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4 years ago

A study showed that 25% of the students drive themselves to school. Based on the suggested probability, in a class of 18 student

Gwar [14]

Answer:

B) 0.283

Step-by-step explanation:

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.

25% of the students drive themselves to school.

This means that $p = 0.25$

Class of 18 students

This means that $n = 18$

What would be the probability that at least 6 students drive themselves to school?

This is

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

In which

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

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

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{18} = 0.006$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{17} = 0.034$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{16} = 0.096$

$P(X = 3) = C_{15,3}.(0.25)^{3}.(0.75)^{15} = 0.17$

$P(X = 4) = C_{15,4}.(0.25)^{4}.(0.75)^{14} = 0.213$

$P(X = 5) = C_{15,5}.(0.25)^{5}.(0.75)^{13} = 0.199$

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.006 + 0.034 + 0.096 + 0.17 + 0.213 + 0.199 = 0.718$

$P(X \geq 6) = 1 - P(X < 6) = 1 - 0.718 = 0.282$

Closest option is B, just a small rounding difference.

4 0

3 years ago