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.
Yep!
Weighing #1: Start off by splitting the pile of 12 coins evenly into two piles, 6 in each pile. Put one pile on each side of the balance. The side that is weighed down has the fake coin it in. Ignore the other 6 coins.
Weighing #2: Now you have 6 coins left. Split the pile evenly again, 3 in each pile. Repeat the same process and put each pile on one side of the balance. The side that is weighed down has your fake coin in it. Ignore the other 3 coins.
Weighing #3: You have 3 coins left. Take two coins, whichever two you like, and weigh them. If they weigh the same, then the one you didn't weigh is the fake one. If one is heavier, then that heavier one is your fake coin.
Answer:
X+y=180, Y=Z, X+Z=180
Step-by-step explanation:
Hope this helps, i remember doing this lesson.
Answer:
x = 31, x = 19
Step-by-step explanation:
