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3 years ago

Ben's $1,000.00 bond earns 7.5% in annual interest. What is the annual interest?

Mathematics

2 answers:

KATRIN_1 [288]3 years ago

6 0

$1000*0.075 = 75$

In short, the answer would be $75.

Hope this helps !

Photon

kvasek [131]3 years ago

4 0

The annual interest is $75.

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What is the answer to x (x+8)=9

marta [7]

Answer:

x=1 or x=−9

Step-by-step explanation:

Let's solve your equation step-by-step.

x(x+8)=9

Step 1: Simplify both sides of the equation.

x2+8x=9

Step 2: Subtract 9 from both sides.

x2+8x−9=9−9

x2+8x−9=0

Step 3: Factor left side of equation.

(x−1)(x+9)=0

Step 4: Set factors equal to 0.

x−1=0 or x+9=0

x=1 or x=−9

8 0

3 years ago

Can you define f(0, 0) = c for some c that extends f(x, y) to be continuous at (0, 0)? If so, for what value of c? If not, expla

Ahat [919]

(i) Yes. Simplify $f(x,y)$ .

$\displaystyle \frac{x^2 - x^2y^2 + y^2}{x^2 + y^2} = 1 - \frac{x^2y^2}{x^2 + y^2}$

Now compute the limit by converting to polar coordinates.

$\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = \lim_{r\to0} \frac{r^4 \cos^2(\theta) \sin^2(\theta)}{r^2} = 0$

This tells us

$\displaystyle \lim_{(x,y)\to(0,0)} f(x,y) = 1$

so we can define $f(0,0)=1$ to make the function continuous at the origin.

Alternatively, we have

$\dfrac{x^2y^2}{x^2+y^2} \le \dfrac{x^4 + 2x^2y^2 + y^4}{x^2 + y^2} = \dfrac{(x^2+y^2)^2}{x^2+y^2} = x^2 + y^2$

and

$\dfrac{x^2y^2}{x^2+y^2} \ge 0 \ge -x^2 - y^2$

Now,

$\displaystyle \lim_{(x,y)\to(0,0)} -(x^2+y^2) = 0$

$\displaystyle \lim_{(x,y)\to(0,0)} (x^2+y^2) = 0$

so by the squeeze theorem,

$\displaystyle 0 \le \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} \le 0 \implies \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = 0$

and $f(x,y)$ approaches 1 as we approach the origin.

(ii) No. Expand the fraction.

$\displaystyle \frac{x^2 + y^3}{xy} = \frac xy + \frac{y^2}x$

$f(0,y)$ and $f(x,0)$ are undefined, so there is no way to make $f(x,y)$ continuous at (0, 0).

(iii) No. Similarly,

$\dfrac{x^2 + y}y = \dfrac{x^2}y + 1$

is undefined when $y=0$ .

5 0

1 year ago

Suppose the parent population has an exponential distribution with a mean of 15 and standard deviation of 12. Use the Central Li

bazaltina [42]

Answer:

The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the population, we have that:

Mean = 15

Standard deviaiton = 12

Sample of 30

By the Central Limit Theorem

Mean 15

Standard deviation $s = \frac{12}{\sqrt{30}} = 2.19$

Approximately normal

The sampling distribution of the sample mean of size 30 will be approximately normal with mean 15 and standard deviation 2.19.

4 0

3 years ago

What is linear expansivity?

katrin [286]

Its a type of thermal expansion and described by fractions that represents the fractional increase length of a thin beam of a material exposed to its temperature increase of one degree celcious,"

4 0

3 years ago

Which point is not on the graph represented by y=x^2+2x-4

Illusion [34]

$(-6,20) \\
x=-6 \\ y=20 \\ \Downarrow \\
20 \stackrel{?}{=} (-6)^2+2 \times (-6)-4 \\
20 \stackrel{?}{=} 36-12-4 \\
20 \stackrel{?}{=} 20 \\
20 = 20 \\
\hbox{the point is on the graph}$

$(-4,4) \\
x=-4 \\ y=4 \\ \Downarrow \\
4 \stackrel{?}{=} (-4)^2+2 \times (-4)-4 \\
4 \stackrel{?}{=} 16-8-4 \\
4 \stackrel{?}{=} 4 \\
4 = 4 \\
\hbox{the point is on the graph}$

Both points are on the graph.

3 0

3 years ago