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3 years ago

5

If you put 200 into a CD that pays 9 percent interest compounded quarterly how long will it take for this money to compound acco

rding to the rule of 72

1 answer:

Kobotan [32]3 years ago

7 0

Well the rule is divide 72 into the interest that is 9 in this case so your answer shall be 8.

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If the slope of a line is -1/4,<br>find x when (x,2) and (1/2,6)

Alecsey [184]

Use slope formula:
m = (y2-y1) / (x2-x1)
-1/4 = 6-2 / 1/2 - x
-1/4 = 4 / 1/2 - x
-1(1/2 - x) = 4(4)
-1/2 + x = 16

x = 16 1/2 or 33/2

7 0

3 years ago

a line with a slope of 1 passes through the point (5,10). what is the equation in this slope-intercept form?

patriot [66]

Answer:

y=x+5

Step-by-step explanation:

So first you need to plug everything into the point-slope formula which is

y-y1=m(x-x1)

m being the slope and y1 and x1 being the point (5,10)

So once you plug it in it should look like this: y-10=1(x-5)

Now you solve it

First you need to distribute 1 in 1(x-5) which is (x-5)

Now you have y-10=x-5

Then you add 10 to both sides which is y=x+5

Your answer is y=x+5

8 0

3 years ago

Your hourly salary increased from $15 to $22 what is the precent increase in your salary?

Elanso [62]

The percentage increase is 46%.

Original number/start = $15

End number = $22

Increase = $22 - $15 = $7

% increase = Increase ÷ Original Number × 100

Substitute in known values

% increase = 7 ÷ 15 × 100

Divide

% increase = 0.46 × 100

Multiply

% increase = 46%

The percentage increase is 46%.

Find out more information about percentage increase here

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8 0

1 year ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

<u>Taylor series</u> expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$

$\text{f}(x)=e^{4x} \implies \text{f}(1)=e^4$

$\text{f}\:'(x)=4e^{4x} \implies \text{f}\:'(1)=4e^4$

$\text{f}\:''(x)=16e^{4x} \implies \text{f}\:''(1)=16e^4$

Substituting the values in the series expansion gives:

$e^{4x}=e^4+4e^4(x-1)+\dfrac{16e^4}{2}(x-1)^2+...$

Factoring out e⁴:

$e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]$

<u>Taylor Series summation notation</u>:

$\displaystyle \text{f}(x)=\sum^{\infty}_{n=0} \dfrac{\text{f}\:^{(n)}(a)}{n!}(x-a)^n$

Therefore:

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

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1 year ago

Use the Normal model N(1156,59) for the weights of steers.

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What weight represents 39th percentile

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3 years ago