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3 years ago

Tim has a $5000 bond with a 4.6% coupon Tim purchased this bond for $5195 what is the yield of the new bond?

Mathematics

1 answer:

NikAS [45]3 years ago

6 0

Answer:

yield of this new bond is 4.42%

Step-by-step explanation:

given data

bond = $5000

coupon rate = 4.6%

purchased bond = $5195

to find out

yield of this new bond

solution

We get here first amount paid to the bond holder

amount paid = 4.6% × $5000

amount paid = $230

and

so Tim earned $230 on a bond that cost her $5195

so yield of this new bond = $\frac{230}{5195} \times 100$

yield of this new bond = 4.42 %

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What is the answer to this question?

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Answer: Choice A) 4, 3, 5, 1, 2

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Explanation:

Since tan(x) = sin(x)/cos(x) is a known and widely used trig identity, this must mean step 4 goes first.

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It turns out that step 3 is the correct statement after step 4. The sin(2x) breaks up into 2sin(x)cos(x), while the cos(2x) breaks up into cos^2(x)-sin^2(x). There are three different variations for the cos(2x) and this just happens to be one of them.

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Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she

Anarel [89]

Answer:

$P(3.25 < \bar X < 4.25)$

And we can use the z score formula given by:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}$

And using the z score we have:

$P(\frac{3.5-4}{0.3}< Z< \frac{4.25-4}{0.3}) = P(-1.667< Z< 0.833)$

And we can use the normal standard distribution table or excel in order to find the probabilities and we got:

$P(-1.667< Z< 0.833)= P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the time it takes her to complete one review, and for this case we know the distribution for X is given by:

$X \sim N(4,1.2)$

Where $\mu=4$ and $\sigma=1.2$

And we select a sample size of n =16. Since the dsitribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by: