2.81+3.72=6.53 Which You Could Estimate To 6.50 or 6.5
The perimeter of each frame would be 24 inches.
24 x 5 = 120
So the total length around all of the frames would be 120 inches.
Answer:
20-45 answer should be d good luck sorrybif im wrong
R^2π=500in^2
r^2=500 in^2/π
r=(500 in^2/π)^(1/2)= 12.62 in
plug it in:
db=3da=3(2(12.62))= 75.69
r of b = 75.69/2=37.85in
A of b = (37.85in)^2π = 8.22 in^2
(a)
The sample space is a set whose elements are all the possible outcomes for the experiment. Since we will extract one of the months of the years, the sample space is the set composed by all the 12 months:
![\Omega = \{ \text{January},\ \text{February},\ldots,\text{December}\}](https://tex.z-dn.net/?f=%20%5COmega%20%3D%20%5C%7B%20%5Ctext%7BJanuary%7D%2C%5C%20%5Ctext%7BFebruary%7D%2C%5Cldots%2C%5Ctext%7BDecember%7D%5C%7D%20)
(b)
An event is a subset of the sample space. Events are often defined by their properties. In this example, the event E is the subset of the sample space defined as
![E = \{ x \in \Omega: x \text{ starts with the letter J}\}](https://tex.z-dn.net/?f=%20E%20%3D%20%5C%7B%20x%20%5Cin%20%5COmega%3A%20x%20%5Ctext%7B%20starts%20with%20the%20letter%20J%7D%5C%7D%20)
So, we have
![E = \{\text{January},\ \text{June},\ \text{July}\}](https://tex.z-dn.net/?f=%20E%20%3D%20%5C%7B%5Ctext%7BJanuary%7D%2C%5C%20%5Ctext%7BJune%7D%2C%5C%20%5Ctext%7BJuly%7D%5C%7D%20)
(c)
If all outcomes have equal probability, then the probability of an event is the ratio bewteen its cardinality, and the cardinality of the whole sample space:
![P(E) = \dfrac{n(E)}{n(\Omega)} = \dfrac{3}{12} = \dfrac{1}{4}](https://tex.z-dn.net/?f=%20P%28E%29%20%3D%20%5Cdfrac%7Bn%28E%29%7D%7Bn%28%5COmega%29%7D%20%3D%20%5Cdfrac%7B3%7D%7B12%7D%20%3D%20%5Cdfrac%7B1%7D%7B4%7D%20)
In words, since there are three months beginning with J out of 12 months, we have a probability of 3 over 12 to pick a month starting with J, which simplifies to 1 over 4.