Q3. Interpolate the following data to estimate the pressure at 1.05 minutes using 2nd order polynomial (By Newton divided difference interpolation) X 1 1.1 1.2 1.3 1.4 1.5 Y 0.14 0.21 0.33 0.54 0.66 0.92
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- Use the table of values you made in part 4 of the example to find the limiting value of the average rate of change in velocity.What is the average velocity, given s= (3t^2+5) ft and t changes from 2s to 3s.Find the z-scores that correspond to the percentage of adult spiders that have carapace lengths between 11 mm and 12 mm. The percentage of adult spiders that have carapace lengths between 11 mm and 12 mm is equal to the area under the standard normal curve between negative 1.72−1.72 and negative 1.18−1.18. (Round to two decimal places as needed.)
- An ecologist hypothesizes that birds with longer wing spans use wider tree branches. The ecologist captured male birds, measured their wing span and other characteristics in millimeters, and then marked and released them. During the ensuing winter, the ecologist repeatedly observed the marked birds as they foraged for food on tree branches. He noted the branch diameter on each occasion, and calculated the average branch diameter for each bird in centimeters. The measurement data are below. What can the ecologist conclude with an α of 0.05 wing span branch diameter 79.380.180.781.579.580.781.180.580.7 1.561.531.391.041.391.021.211.311.21 A) What is the appropriate statistic? Then Compute the statistic B) Compute the appropriate test statistic(s) for H1: ρ > 0 critical value = test statistic= C) Effect SizeThe area beneath the normal density curve is separated into three regions by the values a and b. Set the mean of the normal density curve to 6.4, the standard deviation to 0.7, and the x-value of b to 6.8. First, determine the value of the area to the right of b.If we know the average speed (s1) for an unknown distance, d, and the average speed (s2) of a second journey across the same distance, d, how can we calculate the average speed of the two journeys? {Values of time and distance are not known}
- Suppose the lengths of human pregnancies are normally distributed with u = 266 days and 0 = 16 days. The figure to the right represents the normal curve with u = 266 days and 0 = 16 days. The area to the left of X = 245 is 0.0947. Provide two interpretations of this area.The increase in carbon dioxide (CO2) in the atmosphere is a major cause of global warming. Using data obtained by Charles David Keeling, professor at Scripps Institution of Oceanography, the average amount of CO2 in the atmosphere from 1958 through 2016 is approximated by A(t) = 0.012414t2 + 0.7485t + 313.9 (1 ≤ t ≤ 59) where A(t) is measured in parts per million volume (ppmv) and t in years, with t = 1 corresponding to 1958.† Find the average rate of increase of the average amount of CO2 in the atmosphere from 1958 through 2016. (Round your answer to two decimal places.)If the following specimen is subjected to a tensile test, determine the followings: c- The ultimate tensile strength (ultimate stress), if the maximum load is 120 kN. d- d- The percentage of elongation, if the final length at the fracture point is 55 mm.
- The mean of the differences between before and after measurements is denoted as - D - Xd - Xdifferences - DbarIf we ignore air resistance, a falling body will fall 16t216t2 feet in tt seconds. What is the average velocity between t=2t=2 and t=2.3t=2.3? Round your answer to two decimal places if necessary.The area beneath the normal density curve is separated into three regions by the values a and b. Set the mean of the normal density curve to 4.00, the standard deviation to 0.33, the z-value of a to -3.03 and the z-value of b to be 3.03. determine the area to the left of a, the area between a and b, and the area to the right of b.