Compute the range, interquartile range, and standard deviation for the data presented in prob-lem 3.4. The data are reproduced here. (NOTE: To simplify computations for Q, round off the case numbers that mark the first and third quar-tiles to the nearest whole number.)b. What would happen to the value of the stan dard deviation if you removed Chicago (the city with the highest score) from this distribution and recalculated? Why?

The answer would be 484

Answer: question no. 1 -a. yes, the sides are in the ratio 2 : 5.question no. 2 -b. 59°question no. 3 -c. angles x and y are remote interior angles.question no. 4 -d. 90°, 45° and 45°question no. 5 -b. mn and qrstep-by-step explanation: question no. 1 -rs : xu = 3 : 7.5 = 2 : 5rt : x? = 6.4 : 16 = 2 : 5st : u? = 6 : 15 = 2 : 5as the ratios of all sides are equivalent, the triangles must be congruent and the sides must be proportional.therefore, the answer is: a. yes, the sides are in the ratio 2 : 5.question no. 2 -in the second diagram, angle b is the only letter not shown. therefore, the number must be equivalent to angle b.if we substitute the value of angle b as being 22° in the first diagram, we can now use the angle sum of a triangle to obtain the value of angle a.angle sum of a triangle = 180°therefore: let angle a = aa + 99 + 22 = 180a = 180 - 99 - 22a = 59°therefore, the answer is: b. 59°question no. 3 -angles x and y don't share the vertex of the triangle with the exterior angle w. remote interior angles don't share a vertex of a triangle with an exterior angle. this means that angles x and y are remote interior angles.therefore, the answer is: c. angles x and y are remote interior angles.question no. 4 -the sum of all interior angles of a triangle must be equivalent to 180°.90°, 45° and 45° are the only set of angle measures that are equivalent to 180°.therefore, the answer is: d. 90°, 45° and 45°question no. 5 -the only pair of sides stated in the options which are corresponding are mn and qr.therefore, the answer is: b. mn and qr.

A.5 is your answer to this problem