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kompoz [17]
4 years ago
7

Randy bought a zero coupon bond and a TIPS 5 years ago for $2,500 each. Both had 10 years to maturity. The TIPS's coupon was 2%

while the zero coupon bond will be redeemed for $3,000. What should the maximum real value of his bonds be if he tries to sell them today?
Mathematics
1 answer:
iren2701 [21]4 years ago
5 0
Zero coupon bonds do not earn interest. It is usually sold at a big discount and its redeemable value if beyond its face value will only be redeemed once it reaches maturity.

TIPS stands for Treasury Inflation Protected Securities.

TIPS:
2,500 x 2% x 5 years = 250
2,500 + 250 = 2,750

Zero coupon bond after 5 years: 2,500

Maximum real value of his bonds if he sells them today.

2,750 + 2,500 = 5,250
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Mrrafil [7]

Answer:

We know that in the box there are:

4 twix

3 kit-kat

Then the total number of candy in the box is:

4 +3 = 7

a)

Here we want to find the probability that we draw two twix.

All the candy has the same probability of being drawn from the box.

So, the probability of getting a twix in the first drawn, is equal to the quotient between the number of twix and the total number of candy in the box, this is:

p = 4/7

Now for the second draw, we do the same, but because we have already drawn one twix before, now the number of twix in the box is 3, and the total number of candy in the box is 6.

this time the probability is:

q = 3/6 = 1/2

The joint probability is the product of the individual probabilities, so here we have

P = p*q = (4/7)*(1/2) =  2/7

b) same reasoning than in the previous case:

For the first bar, the probability is:

p = 3/7

for the second bar, the probability is:

q = 2/6 = 1/3

The joint probability is:

P = p*q = (3/7)*(1/3) = 1/7

c) Suppose that first we draw a twix.

The probability we already know that is:

p = 4/7

Now we want another type, so we need to draw a kit-kat, the probability will be equal to the quotient between the remaining kit-kat bars (3) and the total number of candy in the box (6)

q = 3/6

The joint probability is:

P = p*q = (4/7)*(3/6) = 2/7

But, we also have the case where we first draw a kit-kat and after a twix, so we have a permutation of two, then the probability in this case is:

Probability = 2*P = 2*2/7 = 4/7

3 0
3 years ago
Find the sum of the first four terms of the geometric sequence shown below. 4​, 4/ 3​, 4/9​, ...
user100 [1]
It's evident that the first four terms are 4, 4/3, 4/9, and 4/27. So the fourth partial sum of the series is

S_4=4+\dfrac43+\dfrac49+\dfrac4{27}

It's as easy as adding up the fractions, but I bet this is supposed to be an exercise in taking advantage of the fact that the series is geometric and use the well-known formula for computing such a sum.

Multiply the sum by 1/3 and you have

\dfrac13S_4=\dfrac43+\dfrac49+\dfrac4{27}+\dfrac4{81}

Now subtracting this from S_4 gives

S_4-\dfrac13S_4=4-\dfrac4{81}

That is, all the matching terms will cancel. Now solving for S_4, you
have

\dfrac23S_4=4\left(1-\dfrac1{81}\right)
S_4=6\left(1-\dfrac1{81}\right)
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I think it's -11a + 17
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4 years ago
Read 2 more answers
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