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Masteriza [31]
3 years ago
11

You and a friend plan to see 2 movies over the weekend. You can choose from 6 comedy, 2 drama, 4 romance, 1 science fiction, or

3 action movies. You write the movie titles on pieces of paper, place them in a bag, and each randomly select a movie. What is the probability that neither of you selects a comedy? Is this a dependent or independent event? Explain. Please help me..
Mathematics
1 answer:
Debora [2.8K]3 years ago
7 0
There are a total of 16 movies to choose from and 6 of them are comedy. This means that you have a 38% of choosing comedy. But the chances of you not choosing comedy is 62%. And since your friend is also choosing, the chance of neither of you not choosing is cut in half.

So, the final answer of neither of you choosing comedy is 31%

(I don't know whether it's dependent or independent)
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<u>End of year 2 (capital + interest + new deposit)</u>

(1,000*(1.09)+10)*1.09 +10 =

\bf 1,000*(1.09)^2+10(1+1.09)

<u>End of year 3 (capital + interest + new deposit)</u>

\bf (1,000*(1.09)^2+10(1+1.09))(1.09)+10=\\1,000*(1.09)^3+10(1+1.09+1.09^2)

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The sum  

\bf 1+1.09+(1.09)^2+...+(1.09)^{49}

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\bf \frac{(1.09)^{50}-1}{1.09-1}=815.08356

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\bf FV=1,000*(1.09)^{50}+10*815.08356=82,508.35564

The <em>present value PV</em> is

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rounded to the nearest hundredth.

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