Answer:
the question is missing the numbers, so I looked for a similar question:
Suppose you receive $100 at the end of each year for the next three years. a. If the interest rate is 8%, what is the present value of these cash flows? (Answer: $257) b. What is the future value in three years of the present value you computed in (a)? (Answer: $324.61) c. Suppose you deposit the cash flows in a bank account that pays 8% interest per year. What is the balance in the account at the end of each of the next three years (after your deposit is made)? How does the final bank balance compare with your answer in (b)?
a) PV = $100/1.08 + $100/1.08² + $100/1.08³ = $257.71
b) FV = $257.71 x (1 + 8%)³ = $324.64
c) FV = ($100 x 1.08²) + ($100 x 1.08) + $100 = $324.64
it is exactly the same as the answer for (b)
Answer:
The current stock price of Jersey Kids Corp in 2021 is expected to be $39.02.
Explanation:
The current stock price of Jersey Kids Corp in 2021 can be calculated using the formula for the dividend discount model as follows:
P2021 = D2022 / (r - g) ............................ (1)
Where,
P2021 = current stock price in 2021 = ?
D2020 = Annual dividends per share paid in 2020 = $3.00
D2021 = Annual dividends per share paid in 2021 = D2020 * (1 + g) = $3 * (1 + 0.02) = $3.06
D2022 = Annual dividends per share paid in 2022 = D2021 * (1 + g) = $3.06 * (1 + 0.02) = $3.1212
r = required return = 10%. or 0.10
g = growth rate = 2% = 0.02
Substituting the values into equation (2), we have:
P2021 = $3.1212 / (0.10 - 0.02)
P2021 = $3.1212 / 0.08
P2021 = $39.02
Therefore, the current stock price of Jersey Kids Corp in 2021 is expected to be $39.02.
Answer:
a) 
And adding we got:
b) 
And replacing we got:
![P(X \geq 2)= =1-[0.00049 +0.0054] = 0.994](https://tex.z-dn.net/?f=%20P%28X%20%5Cgeq%202%29%3D%20%3D1-%5B0.00049%20%2B0.0054%5D%20%3D%200.994)
c) 
And replacing we got:
Explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest "number of women", on this case we now that:
The probability mass function for the Binomial distribution is given as:
Where (nCx) means combinatory and it's given by this formula:
Part a
For this case we want to find this probability:
And adding we got:
Part b
For this case we want this probability:
And we can use the complement rule and we got:
And replacing we got:
Part c
For this case we want this probability:
And we can use the complement rule and we got:
And replacing we got:
Answer:
False
Bill will have less than Beth
Explanation:
let us compare the amounts accumulated by the two individuals at age 65 to see which is more.
Money accumulated by Bill at age 65:
Since bill saves $2500 a year for 30 years, amount saved at the end of 30 years = 2500 X 30 = $75000.
Money accumulated by Beth at age 65:
Beth saves $2,500 a year from age 25 until age 34: Amount saved = 2500 X 9 = $22500
Invests the money in an account earning ten percent annually for 31 years.
Assuming it was at a simple interest rate,
The interest at the end of the 31st year will be:
Therefore total amount at age 65 = $22, 500 + 69750 = $92,250.
∴ This shows that Bill will have less money than Beth. Primarily due to the fact that he started investing at a lot later time than Beth.
Answer:
The answer is This should be possible in O(m+n) with BFS.
Explanation:
Give us a chance to take your chart G. Complete a BFS on the diagram. Check every one of the hubs in the diagrams as visited as normal with BFS. Rather than adding only hubs to the line in the DFS include hubs in addition to number of incoming ways. On the off chance that a hub that has been visited ought to be included disregard it. On the off chance that you discover a hub again which is as of now present in your line don't include it once more, rather include the checks together. Proliferate the depends on the line while including new hubs when you experience the last hub i.e the goal hub the number that is put away with it is the quantity of briefest ways in the diagram.
