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Nostrana [21]
3 years ago
13

Bonds payable should be reported as a long-term lThe Tony Hawk Skate Park has a 5-year installment loan with monthly payments of

$5,000 including interest. Currently, the first twelve monthly payments have been recorded as a current liability. The VP of Finance, tells board members that he plans to classify the current portion of the installment note as long-term. He explains that lenders will be more willing to let the company borrow and will offer lower interest rate if the company reports fewer current liabilities. iability in the balance sheet at the:
Mathematics
1 answer:
notka56 [123]3 years ago
7 0

Complete question:

The Tony Hawk Skate Park has a 5-year installment loan with monthly payments of $5,000 including interest. Currently, the first twelve monthly payments have been recorded as a current liability. The VP of Finance, tells board members that he plans to classify the current portion of the installment note as long-term. He explains that lenders will be more willing to let the company borrow and will offer lower interest rate if the company reports fewer current liabilities.

Multiple Choice:

a) Classifying the first twelve installment payments as a current liability is not required by GAAP.

b) Reporting the current portion of a long-term note as long-term debt is a misrepresentation of the financial position of the company.

c) The ability to borrow at lower interest rates is more important than being ethical.

d) Presenting investors with a strong current ratio is acceptable at any cost.

Answer:

b) Reporting the current portion of a long-term note as long-term debt is a misrepresentation of the financial position of the company

Step-by-step explanation:

The Tony Hawk Skate Park has a 5-year installment loan with monthly payments of $5,000 including interest, but the first twelve monthly payments have been recorded as a current liability. We are told the VP of Finance, tells board members that he wants to classify the current portion of the installment note as long-term. The current portion of long-term liability would be the amount the company can repay during its operating cycle, the operating cycle in this case is 12 months.

Using the GAAP (Generally Accepted Accounting Principles), since it is paid within 12 months, the current portion of long term liability should be a current liability.

Therefore the VP's plan to classify the current portion of the installment note as long-term debt would be a wrong classification or misrepresentation of the financial position of the company.

The correct option is B.

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chubhunter [2.5K]

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4x<-6-6

Then calculate the difference (-6-6)

4x<-12

Now divide both sides by 4

X<-3

Then you are left with your answer:

x<-3

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The product of two consecutive integers is 72. find the integers
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Answer:

the integers are 8 and 9

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These 20 ages of children on a school bus are in order from least to greatest. 5, 5, 5, 6, 6, 7, 7, 8, 9, 10, 10, 11, 12, 12, 12
umka2103 [35]

Answer:

the interquartile range is the Q3-Q1 where Q3 is 75th percentile and Q1 is the 25th percentile.

so the total count is 20 so 25 percentile is 25% of 20 =5 and 75th percentile is 75% of 20 = 15 and since the number are already arranged in order we just pick the 15th number-5th number   thus, our answer is

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Simplify (9.5)(−2)(−5).
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Historically, there is a 40% chance of having clear sunny skies in Seattle in July. Let's assume that each day is independent fr
Igoryamba

Answer: a)  0.0058

b) 0.0026

Step-by-step explanation:

Given : The probability of having clear sunny skies in Seattle in July : p= 0.40

The number of days spent in Seattle  in July: n= 18

a) Using, Binomial probability formula : P(x)=^nC_xp^x(1-p)^{n-x}

The probability of having clear sunny skies on at least 13 of those days:-

P(x\geq13)=P(13)+P(14)+P(15)+P(16)+P(17)+P(18)\\\\=^{18}C_{13}(0.4)^{13}(0.6)^5+^{18}C_{14}(0.4)^{14}(0.6)^4+^{18}C_{15}(0.4)^{15}(0.6)^3+^{18}C_{16}(0.4)^{16}(0.6)^2+^{18}C_{17}(0.4)^{17}(0.6)^1+^{18}C_{18}(0.4)^{18}(0.6)^0

=\dfrac{18!}{13!5!}(0.4)^{13}(0.6)^5+\dfrac{18!}{14!4!}(0.4)^{14}(0.6)^4+\dfrac{18!}{15!3!}(0.4)^{15}(0.6)^3+\dfrac{18!}{16!2!}(0.4)^{16}(0.6)^2+(18)(0.4)^{17}(0.6)^1+(1)(0.4)^{18}

=0.00447111249474+0.00106455059399+0.000189253438931+0.0000236566798664+0.00000185542587187+0.000000068719476736

=0.00575049735288\approx0.0058

b) On converting binomial to normal distribution, we have

\text{Mean=}\mu=np= 18\times0.40=7.2\\\\\text{Standard deviation}=\sigma=\sqrt{np(1-p)}\\\\=\sqrt{18(0.40)(1-0.40)}=2.07846096908\approx2.08

Let x be the number of days having clear sunny skies in Seattle in July.

Then, using z=\dfrac{x-\mu}{\sigma} we have

z=\dfrac{13-7.2}{2.08}=2.78846153846\approx2.79

P-value = P(x\geq13)=P(z\geq2.79)=1-P(z

=1-0.9973645=0.0026355\approx0.0026

6 0
3 years ago
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