The probability that a randomly selected monthly cell phone bill is more than $127 is given by

$P(X\ \textgreater \ 127)=1-P(X\ \textless \ 127) \\ \\ =1-P\left(z\ \textless \ \frac{127-100}{11.35} \right)=1-P(z\ \textless \ 2.38) \\ \\ =1-0.99132=0.00868$

5 0

3 years ago

Math again yay!...Ew math

Sliva [168]

Answer:

The graph of g(x) is wider.

Step-by-step explanation:

Parent function:

$f(x)=x^2$

New function:

$g(x)=\left(\dfrac{1}{2}x\right)^2=\dfrac{1}{4}x^2$

Transformations:

For a > 0

$f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}$

$f(x)-a \implies f(x) \: \textsf{translated}\:a\:\textsf{units down}$

$\begin{aligned} y =a\:f(x) \implies & f(x) \: \textsf{stretched/compressed vertically by a factor of}\:a\\ & \textsf{If }a > 1 \textsf{ it is stretched by a factor of}\: a\\ & \textsf{If }0 < a < 1 \textsf{ it is compressed by a factor of}\: a\\\end{aligned}$

$\begin{aligned} y=f(ax) \implies & f(x) \: \textsf{stretched/compressed horizontally by a factor of} \: a\\& \textsf{If }a > 1 \textsf{ it is compressed by a factor of}\: a\\ & \textsf{If }0 < a < 1 \textsf{ it is stretched by a factor of}\: a\\\end{aligned}$