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andrey2020 [161]
3 years ago
7

A while ago, Nicole bought shares of stock in a high-tech firm. At that time, the selling price was $60 per share. According to

today's newspaper, a share is currently worth 30% more. What is the current value of one share?
Mathematics
2 answers:
love history [14]3 years ago
8 0

Answer:

Step-by-step explanation:

1. Dividends

2. Capital gains

-The discounted present value of the sum of next period's dividend plus next period's stock price

-The discounted present value of all future dividends

P0 = [(Div1)/(1+R)] + [P1/(1+R)]

where Div 1 = expected dividend paid at year's end

P1 = expected price at year's end

R = discount rate

Marina86 [1]3 years ago
4 0

Answer:

$78

Step-by-step explanation:

A share is worth 30% more, so multiply 30% by 60 to get 30% of 60, which is 18. Then, add 18 to 60 because the share is 30% more, and you get 78.

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Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand
dybincka [34]

Answer: Part a) P(a)=\frac{1}{\binom{r+w}{r}}

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Step-by-step explanation:

The probability is calculated as follows:

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For part a)

Probability that a red ball is drawn in first attempt = P(E_{1})=\frac{r}{r+w}

Probability that a red ball is drawn in second attempt=P(E_{2})=\frac{r-1}{r+w-1}

Probability that a red ball is drawn in third attempt = P(E_{3})=\frac{r-2}{r+w-1}

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = P(E_{i})=\frac{r-i}{r+w-i}

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P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...

Thus P(E)=\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!

Thus our probability becomes

P(E)=\frac{1}{\binom{r+w}{r}}

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at (r+1)^{th} draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals \binom{r}{r-1}

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Thus required probability of case b becomes P(E)+ P(E_{i})

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7 0
4 years ago
Three consecutive odd intergers have a sum of 27. find the integers
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Solve for x:
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The integers are 7, 9 , 11
6 0
3 years ago
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