What’s the equation to this? Help!!

An investment of $5000 earns $6000 simple interest in 12 years. Calculate the interest rate.

adoni [48]

Answer:

10%

Step-by-step explanation:

To solve this problem we can use a modified version of the simple interest formula which is shown below:

$r=\frac{I}{Pt}$

I = interest amount

P = principal amount

t = time (years)

Lets plug in the values:

$r=\frac{6000}{(5000)(12)}$

$r=0.1$

The last step is to convert 0.1 into a percent:

0.1(100) = 10

The interest rate is 10%

4 0

2 years ago

Someone help me pls i need to pass summer school

aleksandr82 [10.1K]

16 to the 0 power 16 to the 0 power 16 to the 0 power 16 to the 0 power 16 to the 0 power

3 0

2 years ago

Read 2 more answers

What is 3/7 +1/5 and how would I get it

Aleks04 [339]

Answer:

22/35

Step-by-step explanation:

3/7 + 1/5

First, find a common denominator.

7*5=35 multiply the numerator by 5 too: 3*5=15

5*7=35 multiply the numerator by 7 too: 1*7=7

15/35 + 7/35

=22/35

Cannot be simplified.

5 0

3 years ago

Read 2 more answers

9. An object that has traveled 400 km in 50 hours, what was the average speed?

Effectus [21]

Answer:

its 8

Step-by-step explanation:

Divide 400÷50

4 0

2 years ago

2) X and Y are jointly continuous with joint pdf

Nady [450]

From what I gather from your latest comments, the PDF is given to be

$f_{X,Y}(x,y)=\begin{cases}cxy&\text{for }0\le x,y \le1\\0&\text{otherwise}\end{cases}$

and in particular, f(x, y) = cxy over the unit square [0, 1]², meaning for 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. (As opposed to the unbounded domain, x ≤ 0 *and* y ≤ 1.)

(a) Find c such that f is a proper density function. This would require

$\displaystyle\int_0^1\int_0^1 cxy\,\mathrm dx\,\mathrm dy=c\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\frac c{2^2}=1\implies \boxed{c=4}$

(b) Get the marginal density of X by integrating the joint density with respect to y :

$f_X(x)=\displaystyle\int_0^1 4xy\,\mathrm dy=(2xy^2)\bigg|_{y=0}^{y=1}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(c) Get the marginal density of Y by integrating with respect to x instead:

$f_Y(y)=\displaystyle\int_0^14xy\,\mathrm dx=\begin{cases}2y&\text{for }0\le y\le1\\0&\text{otherwise}\end{cases}$

(d) The conditional distribution of X given Y can obtained by dividing the joint density by the marginal density of Y (which follows directly from the definition of conditional probability):

$f_{X\mid Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\begin{cases}2x&\text{for }0\le x\le 1\\0&\text{otherwise}\end{cases}$

(e) From the definition of expectation:

$E[X]=\displaystyle\int_0^1\int_0^1 x\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y\,\mathrm dy\right)=\boxed{\frac23}$

$E[Y]=\displaystyle\int_0^1\int_0^1 y\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac23}$

$E[XY]=\displaystyle\int_0^1\int_0^1xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=4\left(\int_0^1x^2\,\mathrm dx\right)\left(\int_0^1y^2\,\mathrm dy\right)=\boxed{\frac49}$

(f) Note that the density of X | Y in part (d) identical to the marginal density of X found in (b), so yes, X and Y are indeed independent.

The result in (e) agrees with this conclusion, since E[XY] = E[X] E[Y] (but keep in mind that this is a property of independent random variables; equality alone does not imply independence.)

8 0

3 years ago