Use the compound interest formulas A = P 1 + r n nt and A = Pe rt to solve. 2) Suppose that you have $8000 to invest. Which investment yields the greater return over 6 years: 6.25% compounded continuously or 6.3% compounded semiannually? 2) A) $8000 invested at 6.3% compounded semiannually over 6 years yields the greater return. B) $8000 invested at 6.25% compounded continuously over 6 years yields the greater return. C) Both investment plans yield the same return.

Answers

Answer 1
Answer:

Answer: A

Compound interest simply defined as the interest added at regular interval. Compound interested can be calculated using

Compound interest = P (1+) ^nt and Pe ^rt

P = Initial balance

r = Annual interest rate

n = Number of times the interest is compounded per year

t =Number of year money is invested

Using

Compound interest = P (1+ ) ^nt

Continuous

P= $ 8000

t = 6

r = 6.25%

=

= 0.0625

n = 1

Compound interest = 8000 (1+) ^1×6

= 8000 (1 + 0.0625) ^6

= 8000 (1.0625) ^ 6

= 8000× 1.4387

= $11,509.6

Semi- annually

P= $ 8000

t = 6

r = 6.3%

=

= 0.063

n = 2

Compound interest = 8000 (1+) ^2×6

= 8000 (1 + 0.063) ^12

= 8000 (1.063) ^12

= 8000× 1.4509

= $11,607.0

Investing $ 8000 semi-annually at 6.3% for 6 years yields greater return

Therefore the answer is (A)


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A principle of $4570 is placed into account that earns 4.5 interest if the interest is compounded annually how much money will be in the account

Answers

Answer:

205.65 interest

Step-by-step explanation:

Answer:

Option D on Edg.

Step-by-step explanation: I took the test and the correct option is $5695.05.

Write an equivalent expression toshow the relationship between
multiplication and addition,
3x5

Answers

I’m not 100% sure what you’re asking for but

5+5+5
3+3+3+3+3
5x3
3x5

A weather service records 28.8 inches of rain over a 12 month period. What is the average rainfall per month for the year?A:
A)
17 inches
B)
2.4 inches
o
2.9 inches
D)
3.1 inches

Answers

Answer:

B) 2.4 inches

Step-by-step explanation:

The total amount of rain is 28.8 inches over 12 months. To find how much rain falls averagely in one month you need to divide the total by the amount of months. 28.8/12 = 2.4

You are the operations manager for an airline and you are considering a higher fare level for passengers in aisle seats. How many randomly selected air passengers must you​ survey? Assume that you want to be 99​% confident that the sample percentage is within 5.5 percentage points of the true population percentage. Complete parts​ (a) and​ (b) below. a. Assume that nothing is known about the percentage of passengers who prefer aisle seats. nequals 549 ​(Round up to the nearest​ integer.) b. Assume that a prior survey suggests that about 34​% of air passengers prefer an aisle seat. nequals nothing ​(Round up to the nearest​ integer.)

Answers

Answer:

a) n=(0.5(1-0.5))/(((0.055)/(2.58))^2)=550.116  

And rounded up we have that n=551

b) n=(0.34(1-0.34))/(((0.055)/(2.58))^2)=493.78  

And rounded up we have that n=494

Step-by-step explanation:

Previous concept

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

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 population proportion have the following distribution

p \sim N(p,\sqrt{(p(1-p))/(n)})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.05. And the critical value would be given by:

z_(\alpha/2)=-2.58, t_(1-\alpha/2)=2.58

Part a

The margin of error for the proportion interval is given by this formula:  

ME=z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}    (a)  

And on this case we have that ME =\pm 0.05 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=(\hat p (1-\hat p))/(((ME)/(z))^2)   (b)  

We can assume that \hat p =0.5 since we don't know prior info. And replacing into equation (b) the values from part a we got:

n=(0.5(1-0.5))/(((0.055)/(2.58))^2)=550.116  

And rounded up we have that n=551

Part b

n=(0.34(1-0.34))/(((0.055)/(2.58))^2)=493.78  

And rounded up we have that n=494

Final answer:

To determine the required sample size for the survey, we can use a sample size formula based on the desired confidence level and margin of error. If nothing is known about the passenger preferences, a sample size of 549 would be needed. If a prior survey suggests a certain proportion, the sample size can be calculated using the known proportion.

Explanation:

In order to determine the number of randomly selected air passengers that must be surveyed, we need to calculate the required sample size for a desired confidence level and margin of error.

a. If nothing is known about the percentage of passengers who prefer aisle seats, we can use a sample size formula given by n = (Z^2 * p * q) / E^2, where Z is the z-score corresponding to the desired confidence level, p and q are the estimated proportions for aisle seat preference and non-aisle seat preference respectively, and E is the desired margin of error. Since a confidence level of 99% and a margin of error of 5.5% are specified, we can round up the sample size to 549.

b. If a prior survey suggests that about 34% of air passengers prefer an aisle seat, we can use the same sample size formula but with the known proportion p = 0.34. We do not have information about the non-aisle seat preference, so we cannot determine the required sample size.

Learn more about Sample size calculation here:

brainly.com/question/34288377

#SPJ11

The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.21 minutes and a standard deviation of 1.90. Find the probability that a randomly selected individual will take less than 6 minutes to select a shoe purchase. Is this outcome unusual?

Answers

Correct answer is: P(x<6) is 0.123 and it is usual.

Solution:-

Given that the time a person takes to decide which shoes to purchase follows normal distribution. Which has mean = 8.21 minutes and standard deviation 1.90

Then probability of individual takes less than 6 minutes is

P(X<6) = P(z<(x-mean)/(standard deviation) )

           = P(z<(6-8.21)/(1.90) )=P(z<-1.163)

           = 0.1230

Typically we say an event with a probability less than 5% is unusual.

But here P(X<6) = 0.123 is greater than 5% hence this is usual.

Given f '(x) = (2 - x)(6 - x), determine the intervals on which f(x) is increasing or decreasing. (2 points)Decreasing (-∞, 2); increasing on (6, ∞)
Decreasing (2, 6); increasing on (-∞, 2) U (6, ∞)
Decreasing (-∞, 2) U (6, ∞); increasing on (2, 6)
Increasing (-∞, -2) U (-6, ∞); increasing on (-2, -6)

Answers

Answer:

Decreasing (2, 6); increasing on (-∞, 2) U (6, ∞)

Step-by-step explanation:

To determine where f(x) is increasing or decreasing, we set f'(x)=0 and check for the intervals.

We see that f'(x)=0 when either x=2 or x=6. Therefore, we'll need to check the intervals (-\infty,2), (2,6), and (6,\infty)

For the interval (\infty,2), we can pick x=0 . This means thatf'(0)=(2-0)(6-0)=(2)(6)=12>0, showing that f(x) increases on the interval (-\infty,2)

For the interval (2,6), we can pick x=4. This means that f'(4)=(2-4)(6-4)=(-2)(2)=-4<0, showing that f(x) decreases on the interval (2,6)

For the interval (6,\infty), we can pick x=7. This means that f'(7)=(2-7)(6-7)=(-5)(-1)=5>0, showing that f(x) increases on the interval (6,\infty)

Therefore, f(x) is increasing on (-\infty,2)\cup(6,\infty) and is decreasing on (2,6).