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Katena32 [7]
3 years ago
14

A credit card had an APR of 11.91% all of last year and compounded interest daily. What was the credit card's effective interest

rate last year?
Mathematics
2 answers:
zysi [14]3 years ago
6 0

Solution:- For credit card, there is interest rate for borrowing money from the bank.  

and that interest rate is termed as yearly rate which is known as annual percentage rate (APR).

Now for the given question APR of a credit card i= 11.91%

and compounded interest daily so, n=365 days

Now for effective interest rate r=(1+\frac{i}{n}^n)-1

r=(1+\frac{0.1191}{365})^{365}-1 \\\Rightarrow\ r=(1+0.0003)^{365}-1\\\\\Rightarrow\ r=(1.0003)^{365}-1=1.1238-1=0.1238

=12.38%

So, the effective interest for last interest is 12.38%

GrogVix [38]3 years ago
5 0

Answer:The correct answer for APEX is 12.64%

Hope this helps :)


Step-by-step explanation:


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1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.
xxTIMURxx [149]

First, rewrite the equation so that <em>y</em> is a function of <em>x</em> :

x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}

(If you were to plot the actual curve, you would have both y=x^{2/3} and y=-x^{2/3}, but one curve is a reflection of the other, so the arc length for 1 ≤ <em>x</em> ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx

We have

y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}

Then in the integral,

\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx

Substitute

u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx

This transforms the integral to

\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du

and computing it is trivial:

\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)

We can simplify this further to

\displaystyle \frac8{27} \left(10\sqrt{10} - \frac{13\sqrt{13}}8\right) = \boxed{\frac{80\sqrt{10}-13\sqrt{13}}{27}}

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3 years ago
What is an equation of the line that passes through the points (6, -3)and (−6,−5)?
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Answer:

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Step-by-step explanation:

First find the slope

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m=(-5 - -3)/( -6 -6)

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The slope intercept form is

y= mx+b where m is the slope and b is the y intercept

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Substitute a point into the equation

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-5 = -1+b

Add 1 to each side

-4 = b

y = 1/6x -4

4 0
3 years ago
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