An article gives arguments for why fracture toughness in concrete specimens should have a Weibull distribution and presents several histograms of data that appear well fit by superimposed Weibull curves. Consider the following sample of size n = 18 observations on toughness for high-strength concrete (consistent with one of the histograms); values of pi = (i − 0.5) 18 are also given. Observation 0.47 0.60 0.65 0.6 0.72 0.74
Pi 0.0278 0.0833 0.1389 0.1944 0.25 0.3056
Observation 0.76 0.79 0.80 0.8 0.82 0.84
Pi 0.3611 0.4167 0.4722 0.5278 0.5833 0.6389
Observation 0.86 0.89 0.91 0.95 .01 1.05
Pi 0.6944 0.75 0.8056 0.8611 0.9167 0.9722
Construct a Weibull probability plot.

Give all possible names for the line shown. calculation tip: lines can be written with either point first.

Since opening night, attendance at play a has increased steadily, while attendance at play b first rose and then fell. equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. on what day(s) was the attendance the same at both plays? what was the attendance? play a: y = 8x + 191 play b: y = -x^2 + 26x + 126 a. the attendance was never the same at both plays. b. the attendance was the same on day 5. the attendance was 231 at both plays on that day. c. the attendance was the same on day 13. the attendance was 295 at both plays on that day. d. the attendance was the same on days 5 and 13. the attendance at both plays on those days was 231 and 295 respectively.

If 2(a^2+b^2)=(a+b)^2 then, > a+b=0, > ab=0, > a=b, > 2a=b