What is the error due to using linear interpolation to estimate the value of sinxsinx at x = \pi/3? your answer should have at
least three significant figures, accurate to within 0.1%. (e.g., 1.23 and 3.33e-8 both have three significant figures.)?
1 answer:
<h3>Answer:</h3>
- using y = x, the error is about 0.1812
- using y = (x -π/4 +1)/√2, the error is about 0.02620
The actual value of sin(π/3) is (√3)/2 ≈ 0.86602540.
If the sine function is approximated by y=x (no error at x = 0), then the error at x=π/3 is ...
... x -sin(x) @ x=π/3
... π/3 -(√3)/2 ≈ 0.18117215 ≈ 0.1812
You know right away this is a bad approximation, because the approximate value is π/3 ≈ 1.04719755, a value greater than 1. The range of the sine function is [-1, 1] so there will be no values greater than 1.
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If the sine function is approximated by y=(x+1-π/4)/√2 (no error at x=π/4), then the error at x=π/3 is ...
... (x+1-π/4)/√2 -sin(x) @ x=π/3
... (π/12 +1)/√2 -(√3)/2 ≈ 0.026201500 ≈ 0.02620
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Answer:
0.13591.
Step-by-step explanation:
We re asked to find the probability of randomly selecting a score between 1 and 2 standard deviations below the mean.
We know that z-score tells us that a data point is how many standard deviation above or below mean.
To solve our given problem, we need to find area between z-score of -2 and -1 that is .
We will use formula to solve our given problem.
Using normal distribution table, we will get:
Therefore, the probability of randomly selecting a score between 1 and 2 standard deviations below the mean would be 0.13591.