Any number divided by 1 is still that number (it stays the same).
So you can rewrite 280 as 280/1, which is a fraction.
With 280/1, you can also get an equivalent fraction by multiplying the numerator and denominator by a number, let's say 2.
So multiplying the numerator 280 by 2 would get 560, the denominator would be 2. So another equivalent fraction would be 560/2
Answer:
The answer would be the one where the graph is between -4 and 4 but not including them. Since it is < it doesn’t include -4 or 4 and since it is less than 4 and absolute value it has to be between them.
These ad agencies must focus on their target audience, which are the students. Hence, they should gather data on the pool that will surely comprise of students. For agency B, social media posting is not a good source pool. It's true that students are very participative and opinionated in social media. However, they can't be sure that these are students. Some parents are in social media, as well. Some are working individuals, and some are out of school youth. Unlike agency A, agency B has to sort out profiles first and identify which ones are students. Hence, agency A will produce a fair sample of the student population because it is unarguably true that everyone in the school enrollment data are students.
The answer is B.
There are 12 such shapes. One can start with two squares, adding squares around the edge and rejecting shapes that have been previously seen. Repeating until you have 5 squares results in 12 distinct shapes.
_____
2 shapes are possible with 3 squares: L, I
5 shapes are possible with 4 squares: L, I, N, T, O
Answer:
79% - almost 4 times out of 5.
The key to this is realizing that the number of games will not always be 5.
If A wins in a sweep - you have 2/3*(2/3)*2/3 percent chance of that happening - 8/27, or 29.63%
If A wins in 4, now we have 2/3*(2/3)*2/3*(1/3)*3 - the 1/3 is the chance that B wins a game. Note - there are only 3 ways B can win a game, not 4. B cannot win Game 4 because Game 4 would not be played in case of a sweep. That is why you cannot use a straight Pascal’s triangle to get your coefficients - the 1–4–6–4–1 is not possible if B cannot win Game 4. Anyway, the math is the same as the above, a 29.63% chance of A winning in 4.
For a 5 game set, A could lose 2 games in 6 possible ways (lose 1&2, 1&3, 1&4, 2&3, 2&4 or 3&4). Again, A cannot lose Game 5 - it would not be played once A wins 3 games. So the odds become 2/3*(2/3)*2/3*(1/3)*(1/3)*6, or 19.75%.
Add them up and you get 79.01%
Step-by-step explanation:
Hope it helps<3
