All categories

2 years ago

What is the difference between a rate that is compounded annually and a rate that is compounded semiannually

Mathematics

1 answer:

8090 [49]2 years ago

3 0

Answer:

the rate compounded semi-annually is compounded twice in a year. thus, this rate is higher than the rate compounded annually which is compounded once in a year

Step-by-step explanation:

The formula for calculating future value:

FV = P (1 + r/m)^mn

FV = Future value

P = Present value

R = interest rate

N = number of years

m = number of compounding

For example, there are two banks

Bank A offers 10% rate with semi-annual compounding

Bank B offers 10% rate with annual compounding.

If you deposit $100, the amount you would have after 2 years in each bank is

A = 100x (1 + 0.1/2)^4 = 121.55

B = 100 x (1 + 0.1)^2 = 121

The interest in bank a is 0.55 higher than that in bank B

You might be interested in

Points B and B' have symmetry with respect to P. Find the coordinates of P when B is (2, 8) and B' is (2, 2). A. (2, 5) B. (0, 5

Alecsey [184]

Answer:

A. (2, 5)

Step-by-step explanation:

If B and B' have symmetry, then P is a midpoint between those points. We can determinate the location of point P by using the midpoint equation, whose vectorial form is:

$P(x,y) = \frac{1}{2}\cdot B(x,y)+\frac{1}{2}\cdot B'(x,y)$ (Eq. 1)

If we know that $B(x,y) = (2,8)$ and $B'(x,y) = (2,2)$ , then the location of P is:

$P(x,y) = \frac{1}{2}\cdot (2,8)+\frac{1}{2}\cdot (2,2)$

$P(x,y) = (1, 4)+(1,1)$

$P(x,y) = (2, 5)$

Which corresponds to option A.

7 0

3 years ago

A piece of bad news about your hotel goes viral. If each person, spreads the news to 5 people within 1 minute, how long will it

Lady_Fox [76]

A 20,000,000

Hope my answer helped

5 0

3 years ago

Read 2 more answers

Find an equation of the tangent line to the curve at the given point. y = x + tan(x) at (pi, pi)

S_A_V [24]

Differentiating :

y' = 1 + sec^2(x),

cosπ=−1

plug in pi = -1 ,for x:

1 + sec^2(pi) = 1 + (-1)^2
= 2

8 0

3 years ago

amber deposits $870 in an account that has a simple interest rate of 8% per year. After 5 years, how much interest will Amber ha

charle [14.2K]

For simple interest, the formula is I = PRT, where I = interest, P = principal borrowed or deposited, R = rate as a decimal, and T = time in years.

Your information:
I = (870)(0.08)(5)
I = 348

Add the interest to the principal for your total balance

348 + 870 = $1218 will be the total balance in the account.

4 0

2 years ago

The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh

mr_godi [17]

Answer:

a) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

b) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(10,0.15)$

Where $\mu=10$ and $\sigma=0.15$

We can calculate the probability of being defective like this:

$P(X$

And we can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

And if we replace we got:

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects.

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

Part c What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0

3 years ago