Answer:
lol, u can still ask the question
Step-by-step explanation:
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.
Answer:
B
Step-by-step explanation
the answer comes to
and 15 is the only answer below 20
1.68 mm per year for 1.5 years could be rewritten as
(2.00-0.32) mm per year for 1.5 yrs.
Multiply: (1.5 yrs)(2.00-0.32)(mm/yr) = 3 mm + 0.48 mm = 3.48 mm
This approach makes use of the distributive property of multiplication.
The answer to this problem is (x,y) = (-4,2)