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If Ana devotes all her time to making fudge, she can make 3 pounds of fudge an hour, and if he devotes all her time to making to
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Answer:
see the attachment
Step-by-step explanation:
We assume that the question is interested in the probability that a randomly chosen class is a Friday class with a lab experiment (2/15). That is somewhat different from the probability that a lab experiment is conducted on a Friday (2/3).
Based on our assumption, we want to create a simulation that includes a 1/5 chance of the day being a Friday, along with a 2/3 chance that the class has a lab experiment on whatever day it is.
That simulation can consist of choosing 1 of 5 differently-colored marbles, and rolling a 6-sided die with 2/3 of the numbers being designated as representing a lab-experiment day. (The marble must be replaced and the marbles stirred for the next trial.) For our purpose, we can designate the yellow marble as "Friday", and numbers greater than 2 as "lab-experiment".
The simulation of 70 different choices of a random class is shown in the attachment.
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<em>Comment on the question</em>
IMO, the use of <em>70 trials</em> is coincidentally the same number as the first <em>70 days</em> of school. The calendar is deterministic, so there will be exactly 14 Fridays in that period. If, in 70 draws, you get 16 yellow marbles, you cannot say, "the probability of a Friday is 16/70." You need to be very careful to properly state the question you're trying to answer.
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.