How do you do this? I'm confused! Thanks for your time and effort!!!
1 answer:
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Question:
James has 14,500g of sand in his sandbox. He brings home another 7,400g of sand from the beach to add to his sandbox. How many kilograms of sand does James have in his sandbox now?
Answer:
The quantity of sandbox James has is 21.9kg
Step-by-step explanation:
Given
Initial quantity of sandbox = 14,500 g
Additional quantity = 7,400 g
Required
Total quantity of sandbox in kg
Tp get the total quantity of sandbox, we simply add in the initial and additional quantity of sandbox together;
This is represented mathematically as follows;
Total = Initial Quantity + Additional Quantity
Total = 14,500 g + 7,400 g
Total = 21,900 g
But the question says the answer should be in kg
1000 g is equivalent to 1 kg
21,900 g will be equivalent to
Hence, the quantity of sandbox James has is 21.9kg
Answer:
The probability that a randomly selected non-defective product is produced by machine B1 is 11.38%.
Step-by-step explanation:
Using Bayes' Theorem
P(A|B) =
where
P(B|A) is probability of event B given event A
P(B|a) is probability of event B not given event A
and P(A), P(B), and P(a) are the probabilities of events A,B, and event A not happening respectively.
For this problem,
Let P(B1) = Probability of machine B1 = 0.3
P(B2) = Probability of machine B2 = 0.2
P(B3) = Probability of machine B3 = 0.5
Let P(D) = Probability of a defective product
P(N) = Probability of a Non-defective product
P(D|B1) be probability of a defective product produced by machine 1 = 0.3 x 0.01 = 0.003
P(D|B2) be probability of a defective product produced by machine 2 = 0.2 x 0.03 = 0.006
P(D|B3) be probability of a defective product produced by machine 3 = 0.5 x 0.02 = 0.010
Likewise,
P(N|B1) be probability of a non-defective product produced by machine 1 = 1 - P(D|B1) = 1 - 0.003 = 0.997
P(N|B2) be probability of a non-defective product produced by machine 2 = 1 - P(D|B2) = 1 - 0.006 = 0.994
P(N|B3) be probability of a non-defective product produced by machine 3 = 1 - P(D|B3) = 1 - 0.010 = 0.990
For the probability of a finished product produced by machine B1 given it's non-defective; represented by P(B1|N)
P(B1|N) = =
= 0.1138
Hence the probability that a non-defective product is produced by machine B1 is 11.38%.