Joseph needed $125 to buy a new pair of jordan's. His neighbor will pay him $6 per hour to walk his dog and his mother gave him
1 answer:
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Answer:
The correct option is (A).
Step-by-step explanation:
Let <em>X</em> = number of orange milk chocolate M&M’s.
The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.
The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.
The event of an milk chocolate M&M being orange is independent of the other candies.
The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.
The expected value of a Binomial random variable is:
Compute the expected number of orange milk chocolate M&M’s in a bag of 55 candies as follows:
It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.
This proportion is not surprising.
This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.
All unusual events have a very low probability, i.e. less than 0.05.
Compute the probability of P (X ≥ 14) as follows:
The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.
This probability is quite larger than 0.05.
Thus, the correct option is (A).
Remark
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.