Question

1. a football team played six games. in those games, the team won by 7 points, lost by 20, won by 8, won by 11, lost by 3, and won by 9. what was the mean amount by which the team won or lost over the six games? a -3 points b 2 points c 3 points d 6 points 20. in digging a hole, the construction crew records the location of the bottom of the hole relative to ground - level. in the first 3 hours, the crew digs at a rate of -2.75 feet each hour. part a how many feet would the crew record after 3 hours? write an expression representing this situation. part b in the next 3 hours, the crew was digging at a rate of -2.5 feet each hour. during the last 2 hours, the crew slowed to a rate of -2.25 feet each hour. what number would they record for the location after the 8 hours? write an expression representing this situation.

Explanation:

Step1: Recall the mean formula

The mean of a set of numbers \(x_1,x_2,\cdots,x_n\) is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\). Here, the points are \(x_1=- 20,x_2 = 8,x_3=11,x_4=-1,x_5 = 9\), and \(n = 5\).

Step2: Calculate the sum of the points

\(\sum_{i=1}^{5}x_i=-20 + 8+11-1 + 9=-20+(8 + 11+9-1)=-20 + 27 = 7\).

Step3: Calculate the mean

\(\bar{x}=\frac{\sum_{i = 1}^{5}x_i}{5}=\frac{7}{5}=1.4\).

For the hole - digging problem:

Part A

Step1: Identify the rate and time

The rate of digging is \(-2.75\) feet per hour and the time is \(3\) hours. The expression for the depth \(d\) is given by the product of the rate \(r\) and the time \(t\). So the expression is \(d=-2.75\times3\).

Part B

Step1: Analyze the different - rate and time intervals

In the first \(3\) hours, the rate \(r_1=-2.75\) feet per hour, so the depth \(d_1=-2.75\times3\). In the next \(3\) hours, the rate \(r_2=-2.5\) feet per hour, so the depth \(d_2=-2.5\times3\). In the last \(2\) hours, the rate \(r_3=-2.75\) feet per hour, so the depth \(d_3=-2.75\times2\). The total - depth \(D\) is \(D=-2.75\times3-2.5\times3-2.75\times2\).

Answer:

For the football - points problem, the mean is \(1.4\).
For Part A of the hole - digging problem, the expression is \(-2.75\times3\).
For Part B of the hole - digging problem, the expression is \(-2.75\times3-2.5\times3-2.75\times2\).