Fill in the blank: A student earned a grade of 80% on a math test that had 20 problems. The student answered ______ questions co
2 answers:
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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:
a) 0.5470 = 54.70% probability that none contain high levels of contamination.
b) 0.3562 = 35.62% probability that exactly one contains high levels of contamination.
c) 0.4530 = 45.30% probability that at least one contains high levels of contamination.
Step-by-step explanation:
For each sample, there are only two possible outcomes. Either they contain high levels of contamination, or they do not. The samples are independent. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
The probability that a lab specimen contains high levels of contamination is 0.14.
This means that
A group of 4 independent samples are checked.
This means that
(a) What is the probability that none contain high levels of contamination?
This is P(X = 0)
0.5470 = 54.70% probability that none contain high levels of contamination.
(b) What is the probability that exactly one contains high levels of contamination?
This is P(X = 1)
0.3562 = 35.62% probability that exactly one contains high levels of contamination.
(c) What is the probability that at least one contains high levels of contamination?
Either none of the samples contain high levels of contamination, or at least one does. The sum of the probabilities of these events is decimal 1. So
We want to find . So
0.4530 = 45.30% probability that at least one contains high levels of contamination.