Given:
The diameter of the circle = 14 cm
To find:
The area of the circle.
Solution:
We have,
Diameter of the circle = 14 cm
We know that the radius of the circle is half of its diameter. So,


The area of a circle is:

Substituting
in the above formula, we get



Therefore, the area of the circle is 154 square cm.
The ratio of the area of the sector (S) to the area of the entire circle (C) is equal to the ratio of the angle subtended to form the sector (As) to the angle for the whole revolution (Ac)
S / C = As / Ac
Substituting the given,
S / 36 m² = 40 degrees / 360 degrees
Solving for S gives, S = 4. Thus, the area of the sector is 4 m².
Answer:
a) -19/60
Step-by-step explanation:
add 7/12 to -19/60 and you get 0.26 repeating.
4/15 is 0.26 repeating.
so yea hope this helped :)
m square +84 is the answer
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:

a) We calculate the probability that you need less than 5 minutes to get up:

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:

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.