How can i simplify 8(3x + 8)
2 answers:
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Answer:
a) 0.2581
b) 0.4148
c) 17
Step-by-step explanation:
For each call, there are only two possible outcomes. Either they are answered in less than 30 seconds. Or they are not. The probabilities for each call 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.
In this problem we have that:
a. If you call 12 times, what is the probability that exactly 9 of your calls are answered within 30 seconds? Round your answer to four decimal places (e.g. 98.7654).
This is when
. So
b. If you call 20 times, what is the probability that at least 16 calls are answered in less than 30 seconds? Round your answer to four decimal places (e.g. 98.7654).
This is when
So
So
c. If you call 22 times, what is the mean number of calls that are answered in less than 30 seconds? Round your answer to the nearest integer.
The expected value of the binomial distribution is:
In this question, we have
So
The closest integer to 16.5 is 17.
It takes 150 seconds for Qian to complete the track.
Let she runs the track in the clockwise direction.
In 60 seconds, she runs =
th of the track.
Since she meets Marissa, who is running in the opposite (counter clockwise) direction in every 60 seconds, Marissa should have completed th of the track.
Chantel meets Marissa in every 300 seconds.
So, in 60 seconds, she advances Marissa by th of the track.
Hence, in 300 seconds, Qian, Marissa and Chantel all meets at the starting point.
ALITER;
Marissa meets Qian in every 60 seconds.
So, Marissa and Qian meet each other in 60th, 120th, 180th, 240th and 300th second.
But, it is given that Chantel meets Marissa in every 300 seconds.
So, in 300 seconds, Qian, Marissa and Chantel meet each other.
