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GalinKa [24]
3 years ago
8

Can anyone help me on this problem? Please it doesn’t make sense to me...

Mathematics
1 answer:
Sav [38]3 years ago
3 0

Answer:

6

Step-by-step explanation:

Cost per item is found by dividing the cost by the number of items. If the woman bought n items for $120, the cost of each item is $120/n. If the woman bought 24 more items, n+24, at the same price, then the cost per item is $120/(n+24). The problem statement tells us this last cost is $16 less than the first cost:

120/(n+24) = (120/n) -16

Multiplying by n(n+24) gives ...

120n = 120(n+24) -16(n)(n+24)

0 = 120·24 -16n^2 -16·24n . . . . . . subtract 120n and collect terms

n^2 +24n -180 = 0 . . . . . . . . . . . . . divide by -16 to make the numbers smaller

(n +30)(n -6) = 0 . . . . . . . . . . . . . . factor the quadratic

The solutions to this are the values of n that make the factors zero: n = -30, n = 6. The negative value of n has no meaning in this context, so n=6 is the solution to the equation.

The woman bought 6 items.

_____

Check

When the woman bought 6 items for $120, she paid $120/6 = $20 for each of them. If she bought 6+24 = 30 items for the same money, she would pay $120/30 = $4 for each item. That amount, $4, is $16 less than the $20 she paid for each item.

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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:

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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:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

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Answer:

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Step-by-step explanation:

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