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.
Independent variable is 5 means x = 5
solve for f(x) when x = 5
f(5) = -2(5) +1
f(5) = -10 + 1
f(5) = -9
Solution (5,-9)
Answer
B. (5,-9)
Answer:
Albar sonic?
Step-by-step explanation:
My grandpa showed me a picture of a car that looked like this. He said it was a Albar sonic.
Answer:
11.9
Step-by-step explanation:
I just know this off the top of my head
The real one is 11.18034
Answer:
- $8000 at 1%
- $2000 at 10%
Step-by-step explanation:
It often works well to let a variable represent the amount invested at the higher rate. Then an equation can be written relating amounts invested to the total interest earned.
__
<h3>setup</h3>
Let x represent the amount invested at 10%. Then 10000-x is the amount invested at 1%. The total interest earned is ...
0.10x +0.01(10000 -x) = 280
<h3>solution</h3>
Simplifying gives ...
0.09x +100 = 280
0.09x = 180 . . . . . . . subtract 100
x = 2000 . . . . . . divide by 0.09
10000 -x = 8000 . . . . amount invested at 1%
<h3>1.</h3>
$8000 should be invested in the 1% account
<h3>2.</h3>
$2000 should be invested in the 10% account
