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3 years ago

By using the table in the handbook, the present value of $12,000 for six years compounded at 6 percent semiannually is

Mathematics

1 answer:

denis23 [38]3 years ago

6 0

Take a look at the attachment to see the solution.
A = future value
P = principal (P = 12,000)
r = interest rate (r=6)
n = time periods (n=12)

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The probability that an individual is left-handed is 0.12. in a class of 39 students, what is the probability of finding five le

Snezhnost [94]

We are given:

p = probability = 0.12
n = total students = 39

x = left handers = 5
u = mean = p* n = 4.68
σ = standard dev = √ ( n*p*(1-p)) = √ ( 39 * 0.12 * 0.88 ) = 2.03

Calculating for the z score:

z = (x – u) / σ
z = (5 – 4.68) / 2.03

z = 0.1576 = 0.16

Using the standard tables for z, the p value is:

p value = 0.5636 = 56.36%

Hence there is a 56.36% chance.

8 0

3 years ago

Read 2 more answers

Explain two methods for solving an equation. Give an example.

svlad2 [7]

For solving system of equations, we can use either substitution where we plug one equation into the other, or elimination where we combine the equations.

- Using elimination, you would to eliminate one variable from both equations, so you automatically would get one equation with one variable!

- Using substitution means you are going to solve one equation for one variable and substitute with its value in the other equation in order to get also an equation with one variable.

Let's take an example ...

y+x=2 and y-2x = 1

Using elimination, we need to subtract these two equation; one from the other...

y+x=2
-
y-2x=1
-----------
0+3x=1

then x=1/3

and then substitute into any equation to get y-value
y+x=2
y+1/3 = 2 >>>>> y=5/3

NOW...Using substitution
y+x=2 and y-2x = 1 >>(y=1+2x)

Plug (y=1+2x) into y+x=2 and solve for x
y+x=2
(1+2x) + x =2
1+3x = 2
3x=1

again (and for sure) x = 1/3

plug in x=1/3 into any of the equations above to get y:
y+x=2
y+1/3=2
y=5/3

DOne !!!!!!

I hope you got the idea

If you still need help, just let me know.

7 0

3 years ago

Determine any asymptotes (Horizontal, vertical or oblique). Find holes, intercepts and state it's domain.

vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

If $x\neq\pm2$ , we can cancel the factors of $x^2-4$ , which makes $x=-2$ and $x=2$ removable discontinuities that appear as holes in the plot of $g(x)$ .

We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

$\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0$

since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

$g(0)=\dfrac{0-4}{0+0-0-4}=1$

so there is one $y$ -intercept at (0, 1).

The domain of $g(x)$ is the set of values that $x$ can take on for which $g(x)$ exists. We've already shown that $x$ can't be -2, 2, or -1, so the domain is the set

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$