Answer:
a) P(x<5)=0.
b) E(X)=15.
c) P(8<x<13)=0.3.
d) P=0.216.
e) P=1.
Step-by-step explanation:
We have the function:
![f(x)=\left \{ {{\frac{1}{10},\, \, \, 10\leq x\leq 20 } \atop {0, \, \, \, \, \, \, otherwise }} \right.](https://tex.z-dn.net/?f=f%28x%29%3D%5Cleft%20%5C%7B%20%7B%7B%5Cfrac%7B1%7D%7B10%7D%2C%5C%2C%20%5C%2C%20%5C%2C%2010%5Cleq%20x%5Cleq%2020%20%7D%20%5Catop%20%7B0%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%5C%2C%20%20otherwise%20%7D%7D%20%5Cright.)
a) We calculate the probability that you need less than 5 minutes to get up:
![P(x](https://tex.z-dn.net/?f=P%28x%3C5%29%3D%5Cint_0%5E5%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%28x%3C5%29%3D%5Cint_0%5E5%200%5C%2C%20dx%5C%5C%5C%5CP%28x%3C5%29%3D0)
Therefore, the probability is P(x<5)=0.
b) It takes us between 10 and 20 minutes to get up. The expected value is to get up in 15 minutes.
E(X)=15.
c) We calculate the probability that you will need between 8 and 13 minutes:
![P(8\leq x\leq 13)=P(10\leqx\leq 13)\\\\P(8\leq x\leq 13)=\int_{10}^{13} f(x)\, dx\\\\P(8\leq x\leq 13)=\int_{10}^{13} \frac{1}{10} \, dx\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot [x]_{10}^{13}\\\\P(8\leq x\leq 13)=\frac{1}{10} \cdot (13-10)\\\\P(8\leq x\leq 13)=\frac{3}{10}\\\\P(8\leq x\leq 13)=0.3](https://tex.z-dn.net/?f=P%288%5Cleq%20x%5Cleq%2013%29%3DP%2810%5Cleqx%5Cleq%2013%29%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cint_%7B10%7D%5E%7B13%7D%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cint_%7B10%7D%5E%7B13%7D%20%5Cfrac%7B1%7D%7B10%7D%20%5C%2C%20dx%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Ccdot%20%5Bx%5D_%7B10%7D%5E%7B13%7D%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Ccdot%20%2813-10%29%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D%5Cfrac%7B3%7D%7B10%7D%5C%5C%5C%5CP%288%5Cleq%20x%5Cleq%2013%29%3D0.3)
Therefore, the probability is P(8<x<13)=0.3.
d) We calculate the probability that you will be late to each of the 9:30am classes next week:
![P(x>14)=\int_{14}^{20} f(x)\, dx\\\\P(x>14)=\int_{14}^{20} \frac{1}{10} \, dx\\\\P(x>14)=\frac{1}{10} [x]_{14}^{20}\\\\P(x>14)=\frac{6}{10}\\\\P(x>14)=0.6](https://tex.z-dn.net/?f=P%28x%3E14%29%3D%5Cint_%7B14%7D%5E%7B20%7D%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%28x%3E14%29%3D%5Cint_%7B14%7D%5E%7B20%7D%20%5Cfrac%7B1%7D%7B10%7D%20%5C%2C%20dx%5C%5C%5C%5CP%28x%3E14%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Bx%5D_%7B14%7D%5E%7B20%7D%5C%5C%5C%5CP%28x%3E14%29%3D%5Cfrac%7B6%7D%7B10%7D%5C%5C%5C%5CP%28x%3E14%29%3D0.6)
You have 9:30am classes three times a week. So, we get:
![P=0.6^3=0.216](https://tex.z-dn.net/?f=P%3D0.6%5E3%3D0.216)
Therefore, the probability is P=0.216.
e) We calculate the probability that you are late to at least one 9am class next week:
![P(x>9.5)=\int_{10}^{20} f(x)\, dx\\\\P(x>9.5)=\int_{10}^{20} \frac{1}{10} \, dx\\\\P(x>9.5)=\frac{1}{10} [x]_{10}^{20}\\\\P(x>9.5)=1](https://tex.z-dn.net/?f=P%28x%3E9.5%29%3D%5Cint_%7B10%7D%5E%7B20%7D%20f%28x%29%5C%2C%20dx%5C%5C%5C%5CP%28x%3E9.5%29%3D%5Cint_%7B10%7D%5E%7B20%7D%20%5Cfrac%7B1%7D%7B10%7D%20%5C%2C%20dx%5C%5C%5C%5CP%28x%3E9.5%29%3D%5Cfrac%7B1%7D%7B10%7D%20%5Bx%5D_%7B10%7D%5E%7B20%7D%5C%5C%5C%5CP%28x%3E9.5%29%3D1)
Therefore, the probability is P=1.
P=2*(l+w)
p=2l+2w
2w=p - 2l
w= (p-2l)/2
Answer:
1/4
Step-by-step explanation:
For the first circle 2 x 2in = 4 to find diameter
4 x 3.14 = 12.56
6in x 2 = 12
12 x 3.14 = 37.68
simplify
12.56/37.68
6.28/18.84
3.14/9.42
So the answer is 1/4 since 3.14 x 2 = 6.28
Also since the radius is 2 times longer