How heavy a load (pounds) is needed to pull apart pieces of douglas fir 4 inches long and 1.5 inches square? here are data from students doing a laboratory exercise:
33,190 31,860 32,590 26,520 33,280
32,320 33,020 32,030 30,460 32,700
23,040 30,930 32,720 33,650 32,340
24,050 30,170 31,300 28,730 31,920
suppose that the strength of pieces of wood like these follows a normal distribution with standard deviation 3000 pounds. give a 90% confidence interval for the mean load required to pull the wood apart.

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23

so what is the problem what are they asking you to solve

step-by-step explanation:

so what is the problem what are they asking you to solve