Step-by-step explanation:
You didn't provide any numbers but in order to find the median of something you order a given list of numbers from least to greatest. After you've done this you find the number located in the middle of the number set. If there are two numbers in the middle then you add them together and divide that number by two.
Hope this helps.
Answer:
About 6 words per second
Step-by-step explanation:
You have to divide 373 words by 60 seconds giving you 6.21666667 which you then have to round down to 6 words per second.
That's false
-7+9=2
and 7+9=18
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:
![f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0](https://tex.z-dn.net/?f=f_%7BX%7D%28x%29%3D%5Clambda%5Ccdot%20e%5E%7B-%5Clambda%20x%7D%3B%5C%20x%3E0%2C%5C%20%5Clambda%3E0)
Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:
![P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx](https://tex.z-dn.net/?f=P%286%5Cleq%20X%5Cleq%209%29%3D%5Cint%5Climits%5E%7B9%7D_%7B6%7D%20%7B%5Clambda%5Ccdot%20e%5E%7B-%5Clambda%20x%7D%7D%20%5C%2C%20dx)
![=\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.