4 friends share 3 oranges equally
2 answers:
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Answer:
Step-by-step explanation:
a. This a binomial probability distribution problem expressed as:
#Given the probability of success, p=0.60 and the number of trials, n=8, the probability of exactly 5 purchases is calculated as:
Hence, the probability of exactly 5 purchases is 0.2787
b. Given n=8 , p=0.60, the probability of more than 5 purchases is calculated as:
Hence, the probability of more than 5 purchases is 0.3154
c. The probability of at most 5 purchases is equivalent to 1-P(x>5) and is calculated as:
#From b above, P(X>5)=0.3154;
Hence, the probability of at most 5 purchases is 0.6846
Honestly I can see an argument for 7 shelves or for 8 shelves.
When you do the calculation;
You get 7.15 shelves. Clearly you cannot have .15 of a shelf so the logician in me wants to say you need 8 shelves!
But if the assignment is aimed estimation skills, (Which is what I assume is being implied by the title of the assignment), then the easier calculation of
is probably what the teacher was hoping for.
You might have been accidentally been calculating
. Remember division is NOT communitive! i.e.
.
From the image;
The first step was to see that 59 can not go into 4..resulting in the zero above the 4, multiplying 0 by 59 and subtracting 0 from 4.
The second step begins by bringing down the 2, seeing that 59 can not go into 42..resulting in the zero above the 2, multiplying 0 by 59 and subtracting 0 from 42.
The third step begins by bringing down the 2, seeing that 59 can go into 422 7 times (I tested this on the right by a short multiplication problem)..resulting in the 7 above the 2, multiplying 7 by 59 and subtracting 413 from 42.
This process continues....
When you do the calculation;
But if the assignment is aimed estimation skills, (Which is what I assume is being implied by the title of the assignment), then the easier calculation of
You might have been accidentally been calculating
From the image;
The first step was to see that 59 can not go into 4..resulting in the zero above the 4, multiplying 0 by 59 and subtracting 0 from 4.
The second step begins by bringing down the 2, seeing that 59 can not go into 42..resulting in the zero above the 2, multiplying 0 by 59 and subtracting 0 from 42.
The third step begins by bringing down the 2, seeing that 59 can go into 422 7 times (I tested this on the right by a short multiplication problem)..resulting in the 7 above the 2, multiplying 7 by 59 and subtracting 413 from 42.
This process continues....