Answer:
You can use my attachment above ^^
180% as a fraction is 180÷100
180÷100= 9÷5 or 1.8 in its simplest form
The answer is 6. If each bagel the man sells costs an extra $2, and the customer bought an amount of bagels so that he ended up paying $12 extra, then he bought 6 bagels (12 / 2 =6)
(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.