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Some of the geometric formulas we take for granted today were first derived by methods that anticipate some of the methods of calculus. The Greek mathematician Archimedes (ca. 287—212; BCE) was particularly inventive, using polygons inscribed within circles to approximate the area of the circle as the number of sides of the
We can estimate the area of a circle by computing the area of an inscribed regular polygon. Think of the regular polygon as being made up of n triangles. By taking the limit as the vertex angle of these mangles goes to zero, you can obtain the area of the circle. To see this, carry out the following steps:
3. If an n-sided regular polygon is inscribed in a circle of radius r, find a relationship between
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CALCULUS,VOLUME 1 (OER)
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- In a crossover trial comparing a new drug to a standard, π denotes the probabilitythat the new one is judged better. It is desired to estimate π and test H0 : π = 0.5against H1 : π = 0.5. In 20 independent observations, the new drug is better eachtime.(a) Find and plot the likelihood function. Give the ML estimate of π (Hint: youmay use the plot function in R)arrow_forwardQ9. If A and B are two events, prove that P(ANB) ≥ 1 − P(Ā) – P(B). [Note: This is a simplified version of the Bonferroni inequality.] -arrow_forwardCan you explain what this analysis means in layman's terms? - We calculated that a target sample size of 3626, which was based on anticipated baseline 90-day mortality of 22% and a noninferiority margin of no more than 4 percentage points, would give the trial 80% power, at a one-sided alpha level of 2.5%, accounting for a maximum of 5% loss to follow-up and for early stopping rules for three interim analyses.-arrow_forward
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- Question 1arrow_forward- Q5. Extend Theorem 5 (P(AUB) = P(A) + P(B) = P(ANB)), proved in class, to three events, A, B and C, by finding an expression for P(AUBUC) in terms of the probabilities of A, B and C, of their pair-wise intersections, and the intersection of all three events. (Hint: Begin by considering AUB as a single event).arrow_forwardCan you help me understand this analysis? A 95.7% confidence interval is shown for the intention-to-treat analysis (accounting for alpha spending in interim analyses), and 95% confidence intervals are shown for the other two analyses. The widths of the confidence intervals have not been adjusted for multiplicity. The dashed line indicates the noninferiority margin of 4 percentage points.arrow_forward
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