Big Y has rice on sale for $1.99 per bag. You bought one bag of rice and 6 cans of black beans. Your total cost was $7.33. How m
2 answers:
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Answer:
1) Probability that all five are good = 0.46
2) P(at most 2 defective) = 0.99
3) Pr(at least 1 defective) = 0.54
Step-by-step explanation:
The total number of bulbs = 30
Number of defective bulbs = 4
Number of good bulbs = 30 - 4 = 26
Number of ways of selecting 5 bulbs from 30 bulbs =
ways
Number of ways of selecting 5 good bulbs from 26 bulbs =
ways
Probability that all five are good = 65780/142506
Probability that all five are good = 0.46
2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)
Pr(no defective) has been calculated above = 0.46
Pr(1 defective) =
Pr(1 defective) = (14950*4)/142506
Pr(1 defective) =0.42
Pr(2 defective) =
Pr(2 defective) = (2600 *6)/142506
Pr(2 defective) = 0.11
P(at most 2 defective) = 0.46 + 0.42 + 0.11
P(at most 2 defective) = 0.99
3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)
Pr(1 defective) =0.42
Pr(2 defective) = 0.11
Pr(3 defective) =
Pr(3 defective) = 0.009
Pr(4 defective) =
Pr(4 defective) = 0.00018
Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018
Pr(at least 1 defective) = 0.54
Answer:
The values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.
Step-by-step explanation:
Note: This question is not complete. The complete question is therefore provided before answering the question as follows:
Determine the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001. f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0
The explanation of the answer is now provided as follows:
Given:
f(x) = e^x ≈ 1 + x + x²/2! + x³/3!, x < 0 …………….. (1)
= (x) = (e^z /4!)x^4
Since the aim is (x) < 0.001, this implies that:
(e^z /4!)x^4 < 0.0001 ………………………………….. (2)
Multiply both sided of equation (2) by (1), we have:
e^4x^4 < 0.024 ……………………….......……………. (4)
Taking 4th root of both sided of equation (4), we have:
|xe^(z/4) < 0.3936 ……………………..........…………(5)
Dividing both sides of equation (5) by e^(z/4) gives us:
|x| < 0.3936 / e^(z/4) ……………….................…… (6)
In equation (6), when z > 0, e^(z/4) > 1. Therefore, we have:
|x| < 0.3936 -----> 0 < x < 0.3936
Therefore, the values of x for which the function can be replaced by the Taylor polynomial if the error cannot exceed 0.001 is 0 < x < 0.3936.