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il63 [147K]
3 years ago
10

Assume you can buy 52 British pounds with 100 Canadian dollars. How much profit can you earn on a triangle arbitrage given the f

ollowing rates if you start out with 100 U.S. dollars?
Business
1 answer:
Nadusha1986 [10]3 years ago
5 0

Answer:

$374.46

Explanation:

Incomplete question. However, I inferred the rates where; CAD/USD=1.35, EUR/USD=1.8305. Thus, using this formular we calculate the profit to be made

=$100 ×(C$1.35 ÷$1) ×(£100 ÷C$52) ×($1.8305 ÷$1)] - $100 = $374.46

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If a company recorded an adjusting entry by debiting Interest Expense for $500 and Interest Payable for $50 in error, then the _
Ilya [14]

If a company recorded an adjusting entry by debiting Interest Expense for $500 and Interest Payable for $50 in error, then the: Adjusted trial balance's debits side will not equal its credits  side.

<h3>Adjusted trial balance</h3>

Based on the information given debiting Interest Expense for $500 was right and debiting Interest Payable for $50 was wrong.

Reason being that  Interest expense is a debit entry because expenses are supposed to be debited while interest payable is a credit entry.

Based on this  the adjusted trial balance's debits side will not equal its credits  side.

Inconclusion If a company recorded an adjusting entry by debiting Interest Expense for $500 and Interest Payable for $50 in error, then the: Adjusted trial balance's debits side will not equal its credits  side.

Learn more about adjusted trial balance here:brainly.com/question/14476257

5 0
2 years ago
A firm has a debt-to-equity ratio of 0.50 and debt equal to $35 million. The firm acquires new equipment with a 3-year operating
ipn [44]

Answer:

($35 million + $12 million) / $70 million = 0.6714

Explanation:

6 0
3 years ago
Using the continuous compounding equation, if someone invested $5,000 at an interest rate of 3.5%, and someone else invested $5,
UNO [17]

Answer:

Therefore after 16.26 unit of time, both accounts have same balance.

The both account have $8,834.43.

Explanation:

Formula for continuous compounding :

P(t)=P_0e^{rt}

P(t)=  value after t time

P_0= Initial principal

r= rate of interest annually

t=length of time.

Given that, someone invested $5,000 at an interest 3.5% and another one  invested $5,250 at an interest 3.2% .

Let after t year the both accounts have same balance.

For the first case,

P= $5,000, r=3.5%=0.035

P(t)=5000e^{0.035t}

For the second case,

P= $5,250, r=3.5%=0.032

P(t)=5250e^{0.032t}

According to the problem,

5000e^{0.035t}=5250e^{0.032t}

\Rightarrow \frac{e^{0.035t}}{e^{0.032t}}=\frac{5250}{5000}

\Rightarrow e^{0.035t-0.032t}=\frac{21}{20}

\Rightarrow e^{0.003t}=\frac{21}{20}

Taking ln both sides

\Rightarrow lne^{0.003t}=ln(\frac{21}{20})

\Rightarrow 0.003t}=ln(\frac{21}{20})

\Rightarrow t}=\frac{ln(\frac{21}{20})}{0.003}

\Rightarrow t= 16.26

Therefore after 16.26 unit of time, both accounts have same balance.

The account balance on that time is

P(16.26)=5000e^{0.035\times 16.26}

              =$8,834.43

The both account have $8,834.43.

7 0
3 years ago
One strength of the team leadership model is ______. a. its complexity b. its application to real-life organizations c. changes
nadezda [96]

Answer:

B. its application to real-life organizations

Explanation:

The answer to this is most likely B because the strength of the team leadership model is application to real-life organizations.

6 0
3 years ago
Theo is depositing $1,300 today in an account with an expected rate of return of 8.1 percent. If he deposits an additional $3,20
SOVA2 [1]

Answer:

$15,699.54

Explanation:

The computation of the account balance after 10 years from today is shown below:

= Future value of amount deposited today × (1 + interest rate)^number of years +   Future value of amount deposited two years × (1 + interest rate)^number of years + Future value of amount deposited three years × (1 + interest rate)^number of years

= $1,300 × (1 + 8.1%)^10 + $3,200 × (1 + 8.1%)^8 + $4,000 × (1 + 8.1%)^7

= $2,832.70 + $5,966.99 + $6,899.85

= $15,699.54

3 0
3 years ago
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