What is the quotient (3x2 + 4x − 15) ÷ (x + 3)? (1 point)
1 answer:
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We can set it up like this, where <em>s </em>is the speed of the canoeist:
To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):
![s(s-5)[\frac{18}{s} + \frac{4}{s-5} = 3] \\ 18(s-5)+4s=3s(s-5) \\ 18s - 90+4s=3 s^{2} -15s](https://tex.z-dn.net/?f=s%28s-5%29%5B%5Cfrac%7B18%7D%7Bs%7D%20%2B%20%5Cfrac%7B4%7D%7Bs-5%7D%20%3D%203%5D%20%5C%5C%2018%28s-5%29%2B4s%3D3s%28s-5%29%20%5C%5C%2018s%20-%2090%2B4s%3D3%20s%5E%7B2%7D%20-15s)
If we rearrange this, we can turn it into a quadratic equation and factor:
Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.
The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.
To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):
If we rearrange this, we can turn it into a quadratic equation and factor:
Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.
The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.
Answer:
P(A) = 0.2
P(B) = 0.25
P(A&B) = 0.05
P(A|B) = 0.2
P(A|B) = P(A) = 0.2
Step-by-step explanation:
P(A) is the probability that the selected student plays soccer.
Then:
P(B) is the probability that the selected student plays basketball.
Then:
P(A and B) is the probability that the selected student plays soccer and basketball:
P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:
P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.