If a fair-number cube with sides numbered from 1 to 6 is tossed 240 times, approximately how many times would the fair-number cu
1 answer:
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Answer:
(a) -3/4
(b) -0.75
(c) -0.75
Step-by-step explanation:
It's a bit hard to tell what constitutes an "iteration" when using the bisection method to approximate a polynomial root. For the purpose here, we'll say one iteration consists of ...
- evaluating the function at the midpoint of the bracketing interval
- choosing a smaller bracketing interval
- identifying the x-value known to be closest to the solution
Thus, the result of the iteration consists of a bracketing interval and the choice of one of the interval's ends as the solution approximation.
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(a) We observe that the graphs intersect in the interval (-1, 0). For the first iteration, we evaluate f(x)-g(x) at x=-1/2. This tells us the solution is in the interval (-1, -1/2). The x-value closest to the root is x=-1/2.
For the second iteration, we evaluate the function f(x)-g(x) at x=-3/4. This tells us the solution is in the interval (-1, -3/4). The x-value closest to the root is x=-3/4.
For the third iteration, we evaluate the function f(x)-g(x) at x=-7/8. This tells us the solution is in the interval (-7/8, -3/4). The x-value closest to the root is x=-3/4.
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(b) The graph tells us the solution is approximately 0.7549. Rounded to 2 decimal places, the solution is approximately 0.75.
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(c) The above solution found after 3 iterations rounded to 2 decimal places is exactly 0.75.
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See the attached table for function values.
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<em>Comment on bisection iteration</em>
Since you cut the interval containing the root in half with each iteration, you gain approximately one decimal place for each 3 iterations. When the function value is very nearly zero at one of the interval endpoints, it can take many more iterations to achieve a better result.
Here, it takes 4 more iterations before an x-value becomes closer to the solution (x≈-97/128). And it takes one more iteration to move the end of the interval away from -3/4. After these 5 more iterations (8 total), the solution is known to lie in the interval (-97/128, -193/256). The corresponding solution approximation is -193/256. It is still only correct to 2 decimal places.
Answer:
The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island
Step-by-step explanation:
From the question, we have;
The distance of the island from the shoreline = 1 mile
The distance the person is staying from the point on the shoreline = 15 mile
The rate at which the visitor runs = 6 mph
The rate at which the visitor swims = 2.5 mph
Let 'x' represent the distance the person runs, we have;
The distance to swim =
The total time, 't', is given as follows;
The minimum value of 't' is found by differentiating with an online tool, as follows;
At the maximum/minimum point, we have;
Simplifying, with a graphing calculator, we get;
-4.72·x² + 142·x - 1,070 = 0
From which we also get x ≈ 15.04 and x ≈ 0.64956
x ≈ 15.04 mile
Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi
Therefore, the distance to run, x ≈ 14.96 mile
Answer:
Choices 1 and 4 are correct.
Step-by-step explanation:
We first need to find what the slope of the line is. That way, we can find out which possible answers are perpendicular to it:
Since we now have the slope, we need the negative reciprocal of it. Remember: if x is the slope, it's negative reciprocal will be . Therefore, if the line's slope is 3, then we need to find answers with a slope of
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The first answer is correct, as you have marked. The second answer, while written a little weirdly, does show the slope as 3, which we know as wrong. The third choice is not correct, however. This equation is written in point-slope form, where. The only variable we have to worry about is m, which, in the third choice, is 3. The fourth answer is correct, which sounds weird at first. Let's put that equation into slope-intercept form:
Equations like these can be real sneaky, so it's important not to jump to conclusions with them.