Hello,
1)
I suppose we must write correcty and corrects the mistakes!
f(x)=x+3
y=x+3
->x=y+3 exchange x and y
->y=x-3 solve for y
f^(-1)(x)=x-3=g(x)
Here are the mistakes:
(fog)(x)=f(g(x))=f(x-3)=(x-3)+3=x and not -x !
(gof)(x)=g(f(x))=g(x+3)=(x+3)-3=x and not -x!!
f(x) and g(x) ARE inverses each another.
2)
f(x)=1/(4x) x≠0
y=1/(4x)
->x=1/(4y)
->y=1/(4x)
g(x)=f^(-1)(x)=1/(4x)=f(x), x≠0
Answer:
Step-by-step explanation:
Find the volume first.
V= l * w * h
l = 2 cm
w = 3 cm
h = 1 cm
V = 2 * 3 * 1 = 6 cm^3
Density = mass / volume
Density = 420 grams / 6 cm^3
Density = 70 grams / cc^3
Kind of large. Check your givens.
Answer:
7.61 miles
Step-by-step explanation:
Given that,
Haley hikes 3 miles north and 7 miles east.
We need to find the shortest distance from the campground to the waterfall. Let the distance is D.
It can be calculated as follows :
So, the shortest distance from the campground to the waterfall is 7.61 miles.
Answer:
The z- score for a value of 3.99 will be 3.38.
Step-by-step explanation:
It is given that the mean of a set of data is -3.82 and its standard deviation is 2.31.
Thus, the value of the z-score=
=
=
=
=
Thus, the z- score for a value of 3.99 will be 3.38.
Answer:
The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.
Step-by-step explanation:
Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.
The random variable <em>X</em> is exponentially distributed with mean 7 minutes.
Then the parameter of the distribution is,
.
The probability density function of <em>X</em> is:
Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:
![=\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148](https://tex.z-dn.net/?f=%3D%5Cint%5Climits%5E%7B9%7D_%7B6%7D%20%7B%5Cfrac%7B1%7D%7B7%7D%5Ccdot%20e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B7%7D%5Ccdot%20%5Cint%5Climits%5E%7B9%7D_%7B6%7D%20%7Be%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C%3D%5B-e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%20x%7D%5D%5E%7B9%7D_%7B6%7D%5C%5C%5C%5C%3De%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%206%7D-e%5E%7B-%5Cfrac%7B1%7D%7B7%7D%20%5Ccdot%209%7D%5C%5C%5C%5C%3D0.424373-0.276453%5C%5C%5C%5C%3D0.14792%5C%5C%5C%5C%5Capprox%200.148)
Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.
