Find m∠UTS, if m∠UTV=x+15, m∠VTS=140∘, and m∠UTS=15x+15.

165

Step-by-step explanation:

First, m<UTS=m<UTV+m<VTS by Angle Addition Postulate.  Then, you substitute all the values that you provided for the angles.  15x+15=x+15+140.  You then solve for x.

15x+15=x+155

14x=140

x=10

You then plug back in 10 for X in the value of m<UTS.  15(10)+15=165

For set p, the average is (3 + 4 + 7 + 8 + 8)/5 = 6. the mean absolute deviation can be solved by taking the absolute values of deviations: |3-6| + |4-6| + |7-6| + |8-6| + |8-6| = 3 + 2 + 1 + 2 + 2 = 10, and dividing by 5 to get mean absolute deviation = 2.for set q, the average is (0 + 0 + 1 + 1 + 3 + 7)/6 = 2. the mean absolute deviation can be solved in a similar way: |0-2| + |0-2| + |1-2| + |1-2| + |3-2| + |7-2| = (2 + 2 + 1 + 1 + 1 + 5)/6 = 2.in set q, the range of 1 mean absolute deviation from the mean is 2 +/- 2, which is from 0 to 4. five of the six points (0, 0, 1, 1, and 3) are within this range.in set p, the range of 2 mean absolute deviation from the mean is 6 +/- 2*2 = 6 +/- 4, which is from 2 to 10. all of the five data points (3, 4, 7, 8 and 8) are within this range.the difference between the means of set p minus set q is 6 - 2 = 4. in this particular case, the difference between the means is twice the mean absolute deviation of each set (both of which have a mean absolute deviation of 2). the comparison of the difference between the means versus each group's mean absolute deviation is a way to see whether there is a statistical difference between the two groups. the larger the difference between the means is compared to each group's mean absolute deviation, the larger the probability that the difference between the groups is statistically significant.