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3 years ago

I need to know the answers to the 3 questions in there

Mathematics

1 answer:

lina2011 [118]3 years ago

3 0

What are the three question?

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A corporate bond has a coupon rate of 5.5 percent, a $1,000 face value, and matures three years from today. The corporation is i

melomori [17]

Answer:

$= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000 }{(1+\frac{15}{100})^3 }$

$=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}$

$=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76$

Step-by-step explanation:

= (probability of entire face value paid*face value+probability of entire face value not paid*percent of face value paid*face value)/(1+discount rate)^years to maturity

probability of entire face value paid = 75%

face value = 1000

probability of entire face value not paid = 25%

percent of face value paid= 60%

discount rate = 15%

years to maturity = 3

$= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000 }{(1+\frac{15}{100})^3 }$

$=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}$

$=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76$

6 0

3 years ago

Which expression is equal to 1/8000 written in scientific notation?

Shkiper50 [21]

Since 1/8000 is 0.000125, we can move the decimal 4 terms to the right.

So that will be 1.25 • 10^-4
Choice B

5 0

3 years ago

for a barbecue mrs.charles bought 12.5 pounds of hamburger meat for $2.89 per pound how much did the hamburger meat cost ? PLEAS

Assoli18 [71]

12.500
x 2.890
 36.125 (rounded up is 36.13)

The hamburger meat costed $36.13.

8 0

3 years ago

Read 2 more answers

What is the total surface area of rectangular prism?

algol13

Answer:

162 [m²].

Step-by-step explanation:

required surface is:

A=(3*7+6*7+3*6)*2=162 [m²].

4 0

3 years ago

Read 2 more answers

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

These are the steps to find the Taylor series for the function $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

To find the first three non-zero terms you need to replace n=3 into the sum

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$

To find the interval on which the series converges you need to use the Ratio Test that says

For the power series centered at x=a

$P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,$

suppose that $\lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.$ . Then

If $R=\infty,$ the the series converges for all x
If $0$ then the series converges for all $|x-a|$
If R=0, the the series converges only for x=a

So we need to evaluate this limit:

$\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |$

Simplifying we have:

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |$

Next we need to evaluate the limit

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}$

-(n+1)(2n+1) is negative when n -> ∞. Therefore $|-(n+1)(2n+1)}|=2n^{2}+3n+1$

You can use this infinity property $\lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty$ when a>0 and n is even. So

$\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty$

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is $( -\infty, \infty )$ .

6 0

3 years ago