<h3>
Answer:</h3>
Debiting salaries Expense $400 and Crediting Salaries payable $400.
<h3>
Explanation:</h3>
We are given;
1 employees earns $ 100 a day
Therefore;
2 employees will earn $ 200 a day
The month ends on Tuesday, but the two employees works on Monday and Tuesday.
- Therefore, the month-end adjusting entry to record will be the amount earned by the two employees on the two days.
Two employees for 2 days = $200/day × 2 days
= $400
- But, salary is an expense, and in the accounts an increase in expense account is debited.
- According to the rule of double entry, an increase in salaries expense decreases the salaries payable. Therefore, we debit salaries expense account and credit salaries payable account.
- Therefore, the month-end adjusting entry to record the salaries earned but unpaid would be;
Debiting salaries Expense $400 and Crediting Salaries payable $400.
Answer:
The answer is 16 years.
Explanation:
The formula for calculating the value of an investment that is compounded annually is given by:
![V(n)=(1+R)^nP](https://tex.z-dn.net/?f=V%28n%29%3D%281%2BR%29%5EnP)
Where:
is the number of years the investment is compounded,
is the annual interest rate,
is the principal investment.
We know the following:
![25000=(1+0.06)^n \times 10000](https://tex.z-dn.net/?f=25000%3D%281%2B0.06%29%5En%20%5Ctimes%2010000)
And we want to clear the value <em>n</em> from the equation.
The problem can be resolved as follows.
<u>First step:</u> divide each member of the equation by
:
![\frac{ 25000}{10000}=(1+0.06)^n \times \frac{ 10000}{10000}](https://tex.z-dn.net/?f=%5Cfrac%7B%2025000%7D%7B10000%7D%3D%281%2B0.06%29%5En%20%5Ctimes%20%5Cfrac%7B%2010000%7D%7B10000%7D)
![2.5=(1.06)^n](https://tex.z-dn.net/?f=2.5%3D%281.06%29%5En)
<u>Second step:</u> apply logarithms to both members of the equation:
![log(2.5)=log (1.06)^n](https://tex.z-dn.net/?f=log%282.5%29%3Dlog%20%281.06%29%5En)
<u>Third step:</u> apply the logarithmic property
in the second member of the equation:
![log(2.5)=n.log (1.06)](https://tex.z-dn.net/?f=log%282.5%29%3Dn.log%20%281.06%29)
Fourth step: divide both members of the equation by ![log1.06](https://tex.z-dn.net/?f=log1.06)
![\frac{log(2.50)}{log (1.06)} =n](https://tex.z-dn.net/?f=%5Cfrac%7Blog%282.50%29%7D%7Blog%20%281.06%29%7D%20%3Dn)
![n= 15.7252](https://tex.z-dn.net/?f=n%3D%2015.7252)
We can round up the number and conclude that it will take 16 years for $10,000 invested today in bonds that pay 6% interest compounded annually, to grow to $25,000.
Answer:
Stock R more beta than Stock S = 4.2%
Explanation:
given data
Stock R beta = 1.8
Stock S beta = 0.75
expected rate of return = 9% = 0.09
risk-free rate = 5% = 0.05
solution
we get here Required Return
Required Return (Re) = risk-free rate + ( expected rate of return - risk-free rate ) beta ...........1
Required Return (Re) = 0.05 + ( 0.09 - 0.05 ) B
Required Return (Re) =
so here
Stock R = 0.05 + ( 0.09 - 0.05 ) 1.8
Stock R = 0.122 = 12.2 %
and
Stock S = 0.05 + ( 0.09 - 0.05 ) 0.75
Stock S = 0.08 = 8%
so here more risky stock is R and here less risky stock is S
Stock R is more beta than the Stock S.
Stock R more beta Stock S = 12.2 % - 8%
Stock R more beta Stock S = 4.2%