1) Johnson's profit is
... net profit = selling price - costs = $155 - 35.00 - 39.17 - 36.43 = $44.40
His net earnings per hour are then
... earnings/hours = $44.40 / (4.5 h) = $9.87/h
2) net profit = selling price - overhead - cost
... net profit = selling price - 0.45 × selling price - cost . . . . substitute for overhead
... net profit = 0.55 × selling price - cost
Given that overhead = 0.45 × selling price, we can find selling price by dividing by 0.45:
... selling price = overhead/0.45
Using this in the profit equation, we have
... net profit = 0.55×(overhead/0.45) - cost
... = (0.55/0.45) × $65.34 - 49.32 . . . . . . substitute given numbers
... net profit = $30.54
First, write out the given datas
Interest I = $11,000
Rate R = 9.5%
Time T = 15 years
Principal P = ?
Next, use the Interest formula below, to find the Principal p.
Next, substitute the value of I, T and R to find the Principal.
Final answer:
P = $7719.3
Answer:
sorry dear but I don't understand your question
P^ = 9/300 = 0.03
H0 = p < 5%
H1 = p > 5%
standard deviation of sample distribution = sqrt[p(1 - p) / n] = sqrt[0.05(1 - 0.05)/300] = sqrt(0.0001583) = 0.01258
test statistics, z = (p^ - p) / standard deviation = (0.03 - 0.05) / 0.01258 = -1.589
P(-1.589) = 1 - P(1.589) = 1 - 0.94402 = 0.05598
Since, p = 0.05598 < significant level of 0.1, we reject the H0.
i.e. There is no sufficient evidence to suggest that the proportion of defective batteries is less than 5%.
The p-value of the test is 0.05598