Answer:
a.) Between 0.5 and 3 seconds.
Step-by-step explanation:
So I just went ahead and graphed this quadratic on Desmos so you could have an idea of what this looks like. A negative quadratic, and we're trying to find when the graph's y-values are greater than 26.
If you look at the graph, you can easily see that the quadratic crosses y = 26 at x-values 0.5 and 3. And, you can see that the quadratic's graph is actually above y = 26 between these two values, 0.5 and 3.
Because we know that the quadratic's graph models the projectile's motion, we can conclude that the projectile will also be above 26 feet between 0.5 and 3 seconds.
So, the answer is a.) between 0.5 and 3 seconds.
D has the commutative property because it just flip flopped the numbers.
Answer:
C. 6 because any answer besides that is too low or too high.
Step-by-step explanation:
110.25 pitches so far- 150 at max. 5.75 pitches per game.
4: 4x5.75 = 23
110.25 + 23 = 133.25
5. 5x5.75 = 28.75
110.25 + 28.75 = 139
6: 6x5.75 = 34.5
110.25 + 34.5 = 144.75
7: 7x5.75 = 40.25
110.25 + 40.25 = 150.50 > which rounds up to 151
Answer:
For every 7 hair bands there will be 4 ribbons
7 hair bands →→ 4 ribbons
14 hair bands →→ 8 ribbons
21 hair bands →→ 12 ribbons
etc.
Answer:
1/6
1/12
1/2
Step-by-step explanation:
There are three possible outcomes in the left spinner, and four possible outcomes in the right spinner. So there are a total of 3×4=12 possible combinations. We can show that by making a grid:
![\left[\begin{array}{cccc}&R&B&G\\R&RR&BR&GR\\B&BR&BB&BG\\P&PR&BP&GP\\Y&RY&BY&GY\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D%26R%26B%26G%5C%5CR%26RR%26BR%26GR%5C%5CB%26BR%26BB%26BG%5C%5CP%26PR%26BP%26GP%5C%5CY%26RY%26BY%26GY%5Cend%7Barray%7D%5Cright%5D)
Of these 12 combinations, 2 show both spinners landing on the same color (RR and BB). So the probability is 2/12 = 1/6.
There is only 1 outcome in which the first spinner lands on R <em>and</em> the second spinner lands on P (PR), so the probability is 1/12.
There are 6 outcomes in which the first spinner lands on R <em>or</em> the second spinner lands on P (RR, BR, PR, RY, BP, GP). So the probability is 6/12 = 1/2.
