How much would $100, growing at 5% per year, be worth after 75 years?a. $3,689.11b. $3,883.27c. $4,077.43d. $4,281.30e. $4,495.3

Haylie ought to consider reallocating her funds elsewhere. Given the current state where her funds are tied up for five years and her investments are at a record low, renewing them wouldn't serve her well. To prepare for her renewal, Haylie should seek various alternatives to ensure she has a strategy for reinvesting her money. She must prioritize finding a bank that offers better terms with a higher interest rate.

Answer:

Zenith Travel Service

Owner's Equity Statement for the year concluded on August 31, 20Y4:

Equity as of September 1, 20Y3 = $456,000

Additional investment                           43,200

Retained Earnings                                 (8,400)

Drawings                                              (21,600)

Equity as of August 31, 20Y4        $469,200

Explanation:

a) Data and Calculations:

Additional investment = $43,200

Personal withdrawals = $21,600

Income Statement for the year ending August 31, 20Y4:

Fees earned                                   $899,600

Office expenses             353,800

Miscellaneous expenses 14,400

Wages expenses           539,800     908,000

Net income/Retained earnings      ($8,400)

b) The statement of owner's equity for Zenith illustrates the modifications within the equity area of its balance sheet during the financial year concluding August 31, 20Y4. Essentially, it exhibits the occurrences impacting Megan Cox's equity from September 1, 20Y3 to August 31, 20Y4.

Response:

$18,000

Clarification:

As per the Internal Revenue Service, the rental property's lifespan is 27.50 years.

Below is the calculation for the highest depreciation amount allowed:

= (Cost of building - value assigned to land - salvage value) ÷ lifespan

= ($600,000 - $104,950 - $0) ÷ 27.50 years

= $495,050 ÷ 27.50 years

= $18,000

Answer:

Probability = 0.3557

Explanation:

Data provided

Young adults age range = 20 to 39

Proportion skipping breakfast p = 0.238

Sample size n = 500

Objective

We aim to determine the probability that more than 122 of Lance's sampled individuals do not eat breakfast regularly.

Solution

Utilizing the Normal Approximation for the Binomial Distribution,

let's define our random variable as x:

x ~ Bin (n,p).............1

Using Normal Approximation gives us:

x ~ Normal Approx (np, npq).................2

This simplifies to:

x ~ (500, 0.238)  

where we know q will be:

q = 1 - p

q = 1 - 0.238

q = 0.762.............3

Hence, we find:

x ~ Normal Approx (119, 90.678)

Now we want to compute P(X > 122).

First, we convert this to Z:

z = ................4

Where the mean is = np

and standard deviation is

For P(X > 122, we evaluate:

............5

This results in P(Z > 0.37).

Thus,

Probability = 1 - P(Z < 0.37)

Next, we'll refer to the z table:

Probability = 1-0.6443

Probability = 0.3557