For this case, the first thing you should keep in mind is:
The solution of the inequation system is given by the shaded region.
Each inequality is met to the right of each graph.
We have then:
For inequation 1:
(2, 6)
(1, -3)
(-2, 5)
For inequation 2:
(2, 6)
(-1,-1)
(-3. 4)
(1, -3)
For the system:
(2, 6)
(1, -3)
Answer:
Y<4
Step-by-step explanation:
Answer:
0.0003
Step-by-step explanation:
Mean=μ=8.21
Standard deviation=σ=2.14
We have to find P(3 randomly monitored call completed in 4 min or less).
P(Xbar≤4)=?
μxbar=μ=8.21
σxbar=σ/√n=2.14/√3=1.2355
Z-score associated with xbar=4
Z=[Xbar-μxbar]/σxbar
Z=[4-8.21]/1.2355
Z=-4.21/1.2355
Z=-3.4075
P(Xbar≤4)=P(Z≤-3.41)
P(Xbar≤4)=P(-∞<Z<0)-P(0<Z<-3.41)
P(Xbar≤4)=0.5-0.4997
P(Xbar≤4)=0.0003
Thus, the probability that three randomly monitored calls will each be completed in 4 minutes or less is 0.0003.