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Ilya [14]
3 years ago
8

The function g(x) = 8|x – 15| - 30 is a transformation of the parent function p(x) = x. Which of the following correctly describ

es the transformation of function p that generates function g?
A p is translated 30 units down, then has a vertical shrink by a factor of 8, and then is translated 15 units right.
B. p has a vertical shrink by a factor of 8, then is translated 15 units down, and then is translated 30 units left
c. p is translated 15 units left, then has a vertical stretch by a factor of 8, and then is translated 30 units up.
D. p is translated 15 units right, then has a vertical stretch by a factor of 8, and then is translated 30 units down.
Mathematics
1 answer:
Anton [14]3 years ago
7 0

Using translation concepts, it is found that the correct option, which represents the transformation made to function g(x) = 8|x - 15| - 30, is given by:

D. p is translated 15 units right, then has a vertical stretch by a factor of 8, and then is translated 30 units down.

The parent function is:

p(x) = |x|

The first transformation was x \rightarrow x - 15, which means that it was shifted 15 units to the right.

Then, the function was <u>multiplied by 8</u>, which means that it was stretched vertically by a factor of 8.

Finally, <u>30 was subtracted from the function</u>, which means that it was shifted down 30 units.

Hence, the correct option is given by:

D. p is translated 15 units right, then has a vertical stretch by a factor of 8, and then is translated 30 units down.

A similar problem is given at brainly.com/question/18405655

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Answer:

See below for answers and explanations

Step-by-step explanation:

<u>Part A</u>

\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478

<u>Part B</u>

P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767

<u>Part C</u>

\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971

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