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3 years ago

14

Sofia invests her money in an account paying 7% interest compounded semiannually. What is the effective annual yield on this acc

ount? Enter your response as a percentage rounded to two decimal places and omit the percentage sign

1 answer:

vfiekz [6]3 years ago

4 0

Answer:

7.12

Step-by-step explanation:

The formula for the effective annual yield is given as:

i = ( 1 + r/m)^m - 1

Where

i = Effective Annual yield

r = interest rate = 7% = 0.07

m= compounding frequency = semi annually = 2

i = ( 1 + 0.07/2)² - 1

i = (1 + 0.035)² - 1

= 1.035² - 1

= 1.071225 - 1

= 0.071225

Converting to percentage

0.071225 × 100

= 7.1225%

Approximately to 2 decimal places = 7.12

Therefore, the annual effective yield = 7.12

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Mark Went to the store and purchase nine pairs of socks and three packages of undershirts for $8.28 each he also purchase a pair

agasfer [191]

Answer:

He paid $2.35 for each pair of socks.

An equation using the variable S is

$77.99 = 9S + (3 × $8.28) + $32

Step-by-step explanation:

Let S be the cost of a pair of socks

From the question, Mark purchase nine pairs of socks, that is, the total cost of the pairs of socks is 9S

He purchase three packages of undershirts for $8.28 each; the total cost of the three packages of shirt will be 3 × $8.28 = $24.84

and he also purchase a pair of pants for $32

Since he spent a total of $77.99, we can write that

$77.99 = 9S + (3 × $8.28) + $32

Now, we can determine S, the cost of a pair of socks

$77.99 = 9S + (3 × $8.28) + $32

$77.99 = 9S + $24.84 + $32

$77.99 = 9S + $56.84

9S = $77.99 - $56.84

9S = $21.15

S = $21.15/9

S = $2.35

Hence, he paid $2.35 for each pair of socks.

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4 years ago

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Andreas93 [3]

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3 years ago

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Usimov [2.4K]

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NNADVOKAT [17]

Answer:

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Step-by-step explanation:

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3 years ago

Find a second solution y2(x) of<br> x^2y"-3xy'+5y=0; y1=x^2cos(lnx)

rosijanka [135]

We can try reduction order and look for a solution $y_2(x)=y_1(x)v(x)$ . Then

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Substituting these into the ODE gives

$x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0$

$x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0$

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which leaves us with an ODE linear in $w(x)=v'(x)$ :

$x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0$

This ODE is separable; divide both sides by the coefficient of $w'(x)$ and separate the variables to get

$w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0$

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$\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx$

Integrate both sides; on the right, substitute $u=\ln x$ so that $\mathrm du=\dfrac{\mathrm dx}x$ .

$\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du$

Now solve for $w(u)$ ,

$\ln|w|=-2\ln(\cos u)-u+C$

$w=e^{-2\ln(\cos u)-u+C}$

$w=Ce^{-u}\sec^2u$

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$w=Ce^{-\ln x}\sec^2(-\ln x)$

$w=C\dfrac{\sec^2(\ln x)}x$

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Substitute $u=\ln x$ again and solve for $v(u)$ :

$v=\displaystyle C_1\int\sec^2u\,\mathrm du$

$v=C_1\tan u+C_2$

then for $v(x)$ ,

$v=C_1\tan(\ln x)+C_2$

So the second solution would be

$y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)$

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$y_1(x)$ already accounts for the second term of the solution above, so we end up with

$\boxed{y_2=x^2\sin(\ln x)}$

as the second independent solution.

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4 years ago