You should calculate( the area of a circle of radius 15 cm ) minus the area of the circle of radius (15 cm - width of the frame). I am not able to read the number, sorry.
That is the surface area of the frame.
Then, if area of circle radius is 100%, the % of face of the clock is:
(100xareaCircle(15cm-widthframe))/areaCircle15cm
Dbd be hehehehe djfjf rnrkfb
Since the sum of the probabilities of all possible outcomes must be 100%, we can deduce the following:
- Cooking in under 20 minutes: 10%
- Cooking between 20 and 30 minutes: 85%
- Cooking in more than 30 minutes: 5%
In fact, the probabilities of cooking in less than 20 or more than 30 sum up to 15%, which means that the remaining outcome (i.e. cooking time between 20 and 30) must complete this probability to 15, and in fact 15+85=100.
That being said, all three answers are simply a combination of these three scenarios: let C be the cooking time, for aesthetic reasons:
![P(C\geq 20) = P(20 \leq C \leq 30)+P(C\geq30) = 85\%+5\%=90\%](https://tex.z-dn.net/?f=P%28C%5Cgeq%2020%29%20%3D%20P%2820%20%5Cleq%20C%20%5Cleq%2030%29%2BP%28C%5Cgeq30%29%20%3D%2085%5C%25%2B5%5C%25%3D90%5C%25)
![P(20 \leq C \leq 30) = 85\%](https://tex.z-dn.net/?f=P%2820%20%5Cleq%20C%20%5Cleq%2030%29%20%3D%2085%5C%25)
![P(C\leq 30) = P(C\leq20)+P(20 \leq C \leq 30) = 10\%+85\%=95\%](https://tex.z-dn.net/?f=P%28C%5Cleq%2030%29%20%3D%20P%28C%5Cleq20%29%2BP%2820%20%5Cleq%20C%20%5Cleq%2030%29%20%3D%2010%5C%25%2B85%5C%25%3D95%5C%25)