3:2
for every 3 ounces of flour, you need 2 ounces of sugar
Table is attached below
From the table we can see that the Lemon cost is equal to 35% of the expenses
Let the expense be x, Lemons cost is 70
Lemon cost = 35% * x
70 = 35% * x ( 35% = 
= 0.35)
70 = 0.35 * x
x = 
x= 200, Expense = $200
Given : Profit percentage is 15%= 
= 0.15
Profit = profit percentage * Expense
Profit = 0.15 * 200 = 30
So profit = $30
Answer:
The triangulation method might not always produce an exact point (other than any measurement errors) because:
The method uses different data sources, researchers, and theories or perspectives.
Step-by-step explanation:
Research Triangulation uses multiple methods, data sources, different investigators, and varied theories for the purpose of developing a more comprehensive understanding of the underlying situations. Triangulation validates the qualitative research by obtaining information from different sources, investigators, viewpoints, or methods.
Lololollolollllollllolloolollllll IM TRYING TO GET POINTS. THX
(a) 0.059582148 probability of exactly 3 defective out of 20
(b) 0.98598125 probability that at least 5 need to be tested to find 2 defective.
(a) For exactly 3 defective computers, we need to find the calculate the probability of 3 defective computers with 17 good computers, and then multiply by the number of ways we could arrange those computers. So
0.05^3 * (1 - 0.05)^(20-3) * 20! / (3!(20-3)!)
= 0.05^3 * 0.95^17 * 20! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18*17! / (3!17!)
= 0.05^3 * 0.95^17 * 20*19*18 / (1*2*3)
= 0.05^3 * 0.95^17 * 20*19*(2*3*3) / (2*3)
= 0.05^3 * 0.95^17 * 20*19*3
= 0.000125* 0.418120335 * 1140
= 0.059582148
(b) For this problem, let's recast the problem into "What's the probability of having only 0 or 1 defective computers out of 4?" After all, if at most 1 defective computers have been found, then a fifth computer would need to be tested in order to attempt to find another defective computer. So the probability of getting 0 defective computers out of 4 is (1-0.05)^4 = 0.95^4 = 0.81450625.
The probability of getting exactly 1 defective computer out of 4 is 0.05*(1-0.05)^3*4!/(1!(4-1)!)
= 0.05*0.95^3*24/(1!3!)
= 0.05*0.857375*24/6
= 0.171475
So the probability of getting only 0 or 1 defective computers out of the 1st 4 is 0.81450625 + 0.171475 = 0.98598125 which is also the probability that at least 5 computers need to be tested.
