By computing the sum of these areas of the smaller rectangles (Lg) and the sum of the areas of the larger rectangles (R8), we obtain better lower and upper estimates for A: 0.5468750 < A < 0.7968750. So one possible answer to the question is to say that the true area lies somewhere between 0.5468750 and 0.7968750. We could obtain better estimates by increasing the number of strips. The table at the left shows the results of similar calculations (with a computer) using n rectangles whose heights are found with left endpoints (Ln) or right endpoints (Rn). In particular, we see by using 50 strips that the area lies between 0.6468 and 0.6868. With 1000 strips, we narrow it down even more: A lies between 0.6656670 and 0.6676670. A good estimate is obtained by averaging these numbers: A =

Q: e z score such that pz is less th

A: we are given that the probability value is 0.7880

Q: find the Z score such that the area to the right of the Z- score is 0.0054.

A:

Q: Find the area under y = 3x2 between x = 0 and x = 1 A B. G. H. 31417/4413 1/2 2.

A:

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A:

Q: 4. For each z-score b X a) z=1.17 %3D X b) z =-1.37

A:

Q: Find the area under the graph of y = 2x² - 2x + 4 on the interval [-2,0].

A: y=2x2−2x+4 on the interval [−2,0]

Q: P(1.15 < Z < 2.77) =

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Q: 10. (8x + 3) dæ

A:

Q: The widths (in meters) of a kidney-shaped swimming pool were measured at 2-meter intervals as…

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Q: he sides of a rectangle measure 5 ± 1 (cm) and 8 ± 2 (cm). How much is your area worth? Consider the…

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A: We can dilate the rectangle with scale factor 1/3.

Q: The region where z>1.92

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A: (b)  X~N(500,85) P(X&lt;x)=0.02

Q: Find Z Z score that corresponds to the cumulative area 0.0059

A: