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IceJOKER [234]
1 year ago
12

Five years ago, Benjamin invested in Parchar Special Effects. He purchased four par value $1,000 bonds from Parchar Special Effe

cts at a market rate of 96.230. Each bond had an interest rate of 7.2%. Benjamin also purchased 200 shares of stock in the same company, each of which cost $19.08 and had a yearly dividend of $2.04. Today, bonds from Parchar Special Effects have a market rate of 104.595, and stock in Parchar Special Effects costs $22.62. If Benjamin liquidates his portfolio and sells all of his investments, which aspect of his investment will have yielded him a greater total profit, and how much greater is it?
a.
The bonds yielded $940.20 more in profits than the stocks.
b.
The bonds yielded $33.00 more in profits than the stocks.
c.
The stocks yielded $373.20 more in profits than the bonds.
d.
The stocks yielded $973.40 more in profits than the bonds.
Mathematics
1 answer:
mart [117]1 year ago
5 0
C if it’s wrong then beat me up
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riadik2000 [5.3K]

Let

S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n

where we assume |r| < 1. Multiplying on both sides by r gives

r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}

and subtracting this from S_n gives

(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}

As n → ∞, the exponential term will converge to 0, and the partial sums S_n will converge to

\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}

Now, we're given

a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a

a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}

We must have |r| < 1 since both sums converge, so

\dfrac{15}a = \dfrac1{1-r}

\dfrac{150}{a^2} = \dfrac1{1-r^2}

Solving for r by substitution, we have

\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)

\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}

Recalling the difference of squares identity, we have

\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}

We've already confirmed r ≠ 1, so we can simplify this to

\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15

It follows that

\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12

and so the sum we want is

ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0
2 years ago
Which factors affect friction between two solid surfaces? Select two options. the weight of the objects the surface area of the
andreev551 [17]

Answer:

a and d

i hope it helps :)))))))))))

5 0
2 years ago
(13 points) asap!! please help!!!!
kenny6666 [7]

Answer:

(-1,4)(1,4)

2

Step-by-step explanation:

6 0
2 years ago
Please help me with these 2 question.
maks197457 [2]

Answer:

Step-by-step explanation:

- sqrt(39) and square root 47

are the limits. The - square root of 39 is smaller than - 6. So the integer to use here is - 6

sqrt (47) = 6.686. Here the square root is larger than the closest integer.

The integer to use is 6

- 6 + 6 = 0

4 0
2 years ago
I need the answer and show work for this question
gavmur [86]

Answer:

f(3) =-12

f(-1) =-12

Step-by-step explanation:

HELLO THERE

f(3)=-2(3)^2+8(3)

f(3)=-36+24

f(3)=-12

----------------------------

f(-1) =2(-1) ^2+8(-1)

f(-1) =-4-8

f(-1) =-12

//Both equal -12//

HAVE A GREAT DAY

6 0
3 years ago
Read 2 more answers
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