Answer:
First on is 0.75
Second is 22/52 (not sure about this one though)
Step-by-step explanation:
For the first one:
P(aUb) = p(a) + p(b) - p(a^b)
0.6 = 0.4 + 0.5 - p(a^b)
p(a^b) = 0.3
P(a/b)= p(a^b)/p(a)
P(a/b)= 0.3/0.4
P(a/b)= 0.75
For the second one:
Total of the cards are 52
Diamonds are 13 out of that 52
Face cards are 12 out of that 52
Diamond OR face cards means p(aUb)
And that equals p(a)+p(b)-p(a^b)
p(a^b) means the intersection between the two, there are three common cards between diamonds and face cards, so p(a^b)=3
13/52 + 12/52 - 3/52= 22/52
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.
Answer:
B
Step-by-step explanation: