Show that A(t)=300−250e0.2−0.02t satisfies the differential equation ⅆAⅆt=6−0.02A with initial condition A(10)=50 .
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Answer:
The likelihood of a battery being produced by Factory A and being defective is 0.012, which corresponds to 1.2%.
Step-by-step explanation:
We have two manufacturers of batteries for mobile phones: Factory A and Factory B. Factory A is responsible for 60% of the total battery production, while Factory B accounts for the remaining 40%.
Out of the batteries made by Factory A, 2% are found to be defective, whereas 4% of those from Factory B are defective.
Let P(A) denote the probability that a battery comes from Factory A, therefore P(A) = 0.60.
The probability that a battery originates from Factory B can be denoted as P(B) = 0.40.
Furthermore, define D as the event indicating that a battery is defective.
Thus, the probability of a defect given that a battery is from Factory A is P(D/A) = 0.02.
For Factory B, the probability of a defect given it is from there is P(D/B) = 0.04.
The probability for a battery being made at Factory A and also being defective can be calculated as follows:
Probability that Factory A produces batteries Probability that a battery from Factory A is defective
= P(A) P(D/A)
= 0.60 0.02 = 0.012 or 1.2%
Conclusively, the probability that a battery is both from Factory A and defective is 1.2%.