Ask question

Ask question

All categories

2 years ago

12

Peter kim wanted to buy a new car.To help finance the purchase he decided to sell his organic markets bond in the secondary mark

et.Peters bond had a par value of $ 10,000 and a coupon of 6 percent.Current interests were 3 percent.What would peters bond sell for?

1 answer:

amid [387]2 years ago

3 0

At a premium, since it pays 6% and market is only 3%.

You might be interested in

D ÷ 20 = 0.81<br> d = ? <br> what is d?

Wittaler [7]

Answer: 16.2

Step-by-step explanation:

i used the calculator

8 0

2 years ago

Please help me i do not understand explain how you answer the following questions.

dlinn [17]

For 6, you could divide 56 kilometers by 8 kilometers to get 7 kilometers. Multiply 7 by 5 to get 35. There would be 35 miles in 56 kilometers

7 0

2 years ago

Brody is in the business of manufacturing phones. He pays a cost of $100 per phone produced for materials and labor and also mus

saw5 [17]

Answer:

C = 100p + 200

Step-by-step explanation:

Because C is the total cost per day, 200 is the y-intercept because it's paid daily.

The 100 is the slope since "he pays a cost of $100 per phone produced)".

8 0

2 years ago

If m 1 = 30° and m = 20°, then m =

Alex

Answer: Then m = 10<span>°</span>

3 0

3 years ago

Required information Skip to question A die (six faces) has the number 1 painted on two of its faces, the number 2 painted on th

grigory [225]

Answer:

The change to the face 3 affects the value of P(Odd Number)

Step-by-step explanation:

Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

The sample space is:

$S = \{1,1,2,2,2,3\}$

And the probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{6}$

$P(1) = \frac{1}{3}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{6}$

$P(2) = \frac{1}{2}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{6}$

P(Odd Number) is then calculated as:

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{6}$

$P(Odd\ Number) = \frac{3}{6}$

$P(Odd\ Number) = \frac{1}{2}$

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

$S = \{1,1,2,2,2,3,3\}$

The probability of each is:

$P(1) = \frac{n(1)}{n(s)}$

$P(1) = \frac{2}{7}$

$P(2) = \frac{n(2)}{n(s)}$

$P(2) = \frac{3}{7}$

$P(3) = \frac{n(3)}{n(s)}$

$P(3) = \frac{1}{7}$

$P(Odd\ Number) = P(1) + P(3)$

$P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}$

Take LCM

$P(Odd\ Number) = \frac{2+1}{7}$

$P(Odd\ Number) = \frac{3}{7}$

Comparing P(Odd Number) before and after

$P(Odd\ Number) = \frac{1}{2}$ --- Before

$P(Odd\ Number) = \frac{3}{7}$ --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0

2 years ago