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3 years ago

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At the end of 2 years, P dollars invested at an interest rate r compounded annually increases to an amount, A dollars, given

by the following formula. Upper A equals Upper P (1 plus r )squared Find the interest rate if $100 increased to $196 in 2 years. Write your answer as a percent.

1 answer:

Valentin [98]3 years ago

5 0

Answer:

40%.

Step-by-step explanation:

We have been given that an amount of $100 compounded annually is increased to $196 in 2 years. We are asked to find the interest rate.

We will use compound interest formula to solve our given problem.

$A=P(1+\frac{r}{n})^{nt}$ , where,

A = Final amount,

P = Principal amount,

r = Annual interest rate in decimal form,

n = Number of times interest is compounded per year,

t = Time in years.

Upon substituting our given values in above formula, we will get:

$196=100(1+\frac{r}{1})^{1*2}$

$196=100(1+r)^{2}$

$\frac{196}{100}=\frac{100(1+r)^{2}}{100}\\\\1.96=(1+r)^2$

$(1+r)^2=1.96$

Take positive square root of both sides:

$\sqrt{(1+r)^2}=\sqrt{1.96}$

$1+r=1.4\\\\1-1+r=1.4-1\\\\r=0.4$

Since interest rate is in decimal, form, so we will convert it into percentage as:

$0.4\times 100\%=40\%$

Therefore, the interest rate was 40%.

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1) Write an equation of a line that<br> is parallel to y = -5x + 2 and<br> passes through (-1, 4).

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Step-by-step explanation:

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The service charge rate are: Help desks $50 per call, Network center $75 per device monitored, Electronic mail $10 per user or per e-mail account, Local voice support $14 per phone extension.

<h3>Service charge rate</h3>

a. Service charge rate

Service charge rate=Budgeted cost/ Budgeted activity base quantity

Help desks:

Service charge rate=$160,000/3,200 calls

Service charge rate=$50 per call

Network center:

Service charge rate=$735,000/9,800 devices

Service charge rate=$75 per device monitored

Electronic mail:

Service charge rate=$100,000/10,000 accounts

Service charge rate=$10 per user or per e-mail account

Local voice support:

Service charge rate=$124,600/8,900 phone extensions

Service charge rate=$14 per phone extension

b. Help desk charge

Help desk charge =(Number of employees times × Percentage of office employees× Percentage with a computer ×Number of calls)× Service charge rate

Help desk charge:

(5,200 employee×25%×96%×1.5)× $50 per call

=$93,600

Network center charge:

[(5,200 employee×25%×96%)+ 600]× $75 per device

=$138,600

Electronic mail:

(5,200 employee×25%×96%×100)×$10 per user or per e-mail account

=$12,480

Local voice support:

(5,200 employee×25%)× $14 per phone extension

=$18,200

Therefore the service charge rate are: Help desks $50 per call, Network center $75 per device monitored, Electronic mail $10 per user or per e-mail account, Local voice support $14 per phone extension.

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2 years ago

Find the distance between the given points: (2, -2) and (-4, 7)

sasho [114]

Answer:

$3\sqrt{13}$

Step-by-step explanation:

Hi there!

We want to find the distance between the points (2, -2) and (-4, 7).

To do that, we can use the distance formula.

The distance formula is given as $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , where $(x_1, y_1)$ and $(x_2, y_2)$ are points

We have everything needed to find the distance, but let's label the values of the points to avoid any confusion

$x_1=2\\y_1=-2\\x_2=-4\\y_2=7$

Now substitute those values into the formula and solve

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\sqrt{(-4-2)^2+(7--2)^2}$

Simplify

$\sqrt{(-4-2)^2+(7+2)^2}$

$\sqrt{(-6)^2+(9)^2}$

Square the numbers under the radical

$\sqrt{36+81}$

Add the numbers under the radical together

$\sqrt{117}$

Simplify the square root

$3\sqrt{13}$

Hope this helps!

6 0

3 years ago

If a and b are positive numbers, find the maximum value of f(x) = x^a(2 − x)^b on the interval 0 ≤ x ≤ 2.

Ad libitum [116K]

Answer:

The maximum value of f(x) occurs at:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Step-by-step explanation:

Answer:

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0$

And we want to find the maximum value of f(x) on the interval [0, 2].

First, let's evaluate the endpoints of the interval:

$\displaystyle f(0) = (0)^a(2-(0))^b = 0$

And:

$\displaystyle f(2) = (2)^a(2-(2))^b = 0$

Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:

$\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]$

By the Product Rule:

$\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}$

Set the derivative equal to zero and solve for <em>x: </em>

$\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]$

By the Zero Product Property:

$\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0$

The solutions to the first equation are <em>x</em> = 0 and <em>x</em> = 2.

First, for the second equation, note that it is undefined when <em>x</em> = 0 and <em>x</em> = 2.

To solve for <em>x</em>, we can multiply both sides by the denominators.

$\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))$

Simplify:

$\displaystyle a(2-x) - b(x) = 0$

And solve for <em>x: </em>

$\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\ \frac{2a}{a+b} &= x \end{aligned}$

So, our critical points are:

$\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}$

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b$

This can be simplified to:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Since <em>a</em> and <em>b</em> > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.

To confirm that this is indeed a maximum, we can select values to test. Let <em>a</em> = 2 and <em>b</em> = 3. Then:

$\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)$

The critical point will be at:

$\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8$

Testing <em>x</em> = 0.5 and <em>x</em> = 1 yields that:

$\displaystyle f'(0.5) >0\text{ and } f'(1)$

Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.

Therefore, the maximum value of f(x) occurs at:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

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3 years ago