If you shift the quadratic parent function, f(x)=x^2, right 12 units, what is the equation of the new function?

Mathematics

2 answers:

aliya0001 [1]3 years ago

8 0

Answer:

G(x)=(x-12)^2

Step-by-step explanation:

Kobotan [32]3 years ago

7 0

Answer:

f(x)=(x-12)^2

Step-by-step explanation:

We know that the function f(x+m) equals to the function f(x) translated 'm' times to the left. Therefore, to translate the function f(x) = x^2 twelve units to the right, we have to:

f(x-12) = (x-12)^2

Therefore, the new function is: f(x)=(x-12)^2

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Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b

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a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$