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3 years ago

(1). There are always 4 bars of chocolates,

Mathematics

1 answer:

liq [111]3 years ago

6 0

Answer:

The smallest number of chocolates in the box is 124.

Step-by-step explanation:

There are always 4 bars of chocolates left in the box when they are share equally among 8, 10, or 12 children.

This means that the smallest number of chocolates in the box is 4 added to the least common multiple of 8, 10 and 12.

LCM(8,10,12).

We factore them by prime factors until all are 1. So

8 - 10 - 12|2

4 - 5 - 6|2

2 - 5 - 3|2

1 - 5 - 3|3

1 - 5 - 1|5

1 - 1 - 1

lcm(8,10,12) = 2*2*2*3*5 = 8*15 = 120

Adding 4: 120 + 4 = 124

The smallest number of chocolates in the box is 124.

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Answer:

$(0.6)^{n}+n(0.4)(0.6)^{n-1}+\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

Step-by-step explanation:

We will use the binomial distribution. Let X be the random variable representing the no. of boxes Hannah buys before betting a prize.

Our success is winning the prize, p =40/100 = 0.4

Then failure q = 1-0.4 = 0.6

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P(X ≤ 3) = P(X=0) +P(X=1)+P(X=2)+P(X=3)

= $\binom{n}{0}p^{0}q^{n-0}$ + $\binom{n}{1}p^{1}q^{n-1}$ + $\binom{n}{2}p^{2}q^{n-2}$ + $\binom{n}{3}p^{3}q^{n-3}$

= $q^{n}+npq^{n-1}$ + $\frac{n(n-1)(p)^{2}q^{n-2}}{2}$ + $\frac{n(n-1)(n-2)p^{3}q^{n-3}}{6}$

= $(0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}$

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