NOTE: Your question isn't clear, Johnson. Would you mind checking it and writing it in a way you can be better helped?
Meanwhile, I hope these explanation below helps.
Answer and Explanation:
Two goods are said to be complementary goods if an increase in the price of a particular one leads to a commensurate decrease in the demand that buyers placed for the other one.
A good is said to be a normal good if the reason for an increase in demand is due to an increase in the income of the buyers.
A good is said to be an inferior good if there is a decrease in demand even though the buyers have experienced increase in their income.
Answer:
Explanation: Dr Cr
1)
Allowance for doubtful account
3%*3610000 108300
Bad debt expense 108300
2)Allowance for doubtful account
2%*(1285070+3610000) 146,852.10
Bad debt expense 146,852.10
3)Allowance for doubtful account
1093830*6% 65,629.80
Bad debt expense 65,629.80
Answer:
It means exchange for good or service without using any money
Answer:
$587.79
Explanation:
Data provided in the question
Amount paid in three years = $700
Discount rate in the first year = 5%
Discount rate in the second year = 6%
Discount rate in the third year = 7%
So by considering the above information, the present value is
= (Amount paid in three years) ÷ (1 + Discount rate in the first year × 1 + Discount rate in the second year × 1 + Discount rate in the third year)
= ($700) ÷ (1 + 0.05 × 1 + 0.06 × 1 + 0.07)
= ($700) ÷ (1.05 × 1.06 × 1.07)
= $700 ÷ 1.19091
= $587.79
Answer:
The answer is below
Explanation:
The marginal revenue R'(t) = 
and the marginal cost C'(t) = 140 - 0.3t.
The total profit is the difference between the total revenue and total cost of a product, it is given by:
Profit = Revenue - Cost
P(T) = R(T) - C(T)
P(T) = ∫ R'(T) - C'(T)
Hence the total profit from 0 to 5 days is given as
![P(T) = \int\limits^0_5 {(R'(T)-C'(T))} \, dt= \int\limits^0_5 {(100e^t-(140-0.3t))} \, dt\\ \\P(T)= \int\limits^0_5 {(100e^t-140+0.3t))} \, dt\\\\P(T)= \int\limits^0_5 {100e^t} \, dt- \int\limits^0_5 {140} \, dt+ \int\limits^0_5 {0.3t} \, dt\\\\P(T)=100\int\limits^0_5 {e^t} \, dt- 140\int\limits^0_5 {1} \, dt+0.3 \int\limits^0_5 {t} \, dt\\\\P(T)=100[e^t]_0^5-140[t]_0^5+0.3[\frac{t^2}{2} ]_0^5\\\\P(T)=100(147.41)-140(5)+0.3(12.5)=14741-700+3.75\\\\P(T)=14045](https://tex.z-dn.net/?f=P%28T%29%20%3D%20%5Cint%5Climits%5E0_5%20%7B%28R%27%28T%29-C%27%28T%29%29%7D%20%5C%2C%20dt%3D%20%5Cint%5Climits%5E0_5%20%7B%28100e%5Et-%28140-0.3t%29%29%7D%20%5C%2C%20dt%5C%5C%20%5C%5CP%28T%29%3D%20%5Cint%5Climits%5E0_5%20%7B%28100e%5Et-140%2B0.3t%29%29%7D%20%5C%2C%20dt%5C%5C%5C%5CP%28T%29%3D%20%5Cint%5Climits%5E0_5%20%7B100e%5Et%7D%20%5C%2C%20dt-%20%5Cint%5Climits%5E0_5%20%7B140%7D%20%5C%2C%20dt%2B%20%5Cint%5Climits%5E0_5%20%7B0.3t%7D%20%5C%2C%20dt%5C%5C%5C%5CP%28T%29%3D100%5Cint%5Climits%5E0_5%20%7Be%5Et%7D%20%5C%2C%20dt-%20140%5Cint%5Climits%5E0_5%20%7B1%7D%20%5C%2C%20dt%2B0.3%20%5Cint%5Climits%5E0_5%20%7Bt%7D%20%5C%2C%20dt%5C%5C%5C%5CP%28T%29%3D100%5Be%5Et%5D_0%5E5-140%5Bt%5D_0%5E5%2B0.3%5B%5Cfrac%7Bt%5E2%7D%7B2%7D%20%5D_0%5E5%5C%5C%5C%5CP%28T%29%3D100%28147.41%29-140%285%29%2B0.3%2812.5%29%3D14741-700%2B3.75%5C%5C%5C%5CP%28T%29%3D14045)
