D=426-48(21)
I got 18 days.
So, D=18
Hope this helps! :)
So we want to know when the cost of each is the same:
2.5m=7.5+m subtract m from both sides
1.5m=7.5 divide both sides by 1.5
m=5
So renting 5 movies costs the same from either store. ($12.50)
It's hard to tell what the independent variable is here. Since time is often plotted on the horizontal axis, that's what has been done here.
Answer: $40000
Step-by-step explanation:
200000(0.2) = $40000
a.
![P({A_1}^c)=1-P(A_1)=\boxed{0.86}](https://tex.z-dn.net/?f=P%28%7BA_1%7D%5Ec%29%3D1-P%28A_1%29%3D%5Cboxed%7B0.86%7D)
b.
![P(A_1\cap A_2)=P(A_1)+P(A_2)-P(A_1\cup A_2)=\boxed{0.07}](https://tex.z-dn.net/?f=P%28A_1%5Ccap%20A_2%29%3DP%28A_1%29%2BP%28A_2%29-P%28A_1%5Ccup%20A_2%29%3D%5Cboxed%7B0.07%7D)
c. By the law of total probability,
![P({A_3}^c)=P((A_1\cap A_2)\cap{A_3}^c)+P((A_1\cap A_2)^c\cap{A_3}^c)](https://tex.z-dn.net/?f=P%28%7BA_3%7D%5Ec%29%3DP%28%28A_1%5Ccap%20A_2%29%5Ccap%7BA_3%7D%5Ec%29%2BP%28%28A_1%5Ccap%20A_2%29%5Ec%5Ccap%7BA_3%7D%5Ec%29)
According to the inclusion/exclusion principle,
![P((A_1\cap A_2)^c\cap{A_3}^c)=P((A_1\cap A_2)^c)+P({A_3}^c)-P((A_1\cap A_2)^c\cup{A_3}^c)](https://tex.z-dn.net/?f=P%28%28A_1%5Ccap%20A_2%29%5Ec%5Ccap%7BA_3%7D%5Ec%29%3DP%28%28A_1%5Ccap%20A_2%29%5Ec%29%2BP%28%7BA_3%7D%5Ec%29-P%28%28A_1%5Ccap%20A_2%29%5Ec%5Ccup%7BA_3%7D%5Ec%29)
but by DeMorgan's law,
![(A_1\cap A_2)^c\cup{A_3}^c={A_1}^c\cup{A_2}^c\cup{A_3}^c=(A_1\cap A_2\cap A_3)^c](https://tex.z-dn.net/?f=%28A_1%5Ccap%20A_2%29%5Ec%5Ccup%7BA_3%7D%5Ec%3D%7BA_1%7D%5Ec%5Ccup%7BA_2%7D%5Ec%5Ccup%7BA_3%7D%5Ec%3D%28A_1%5Ccap%20A_2%5Ccap%20A_3%29%5Ec)
So
![P((A_1\cap A_2)^c\cap{A_3}^c)=(1-P(A_1\cap A_2))+(1-P(A_3))-(1-P(A_1\cap A_2\cap A_3))](https://tex.z-dn.net/?f=P%28%28A_1%5Ccap%20A_2%29%5Ec%5Ccap%7BA_3%7D%5Ec%29%3D%281-P%28A_1%5Ccap%20A_2%29%29%2B%281-P%28A_3%29%29-%281-P%28A_1%5Ccap%20A_2%5Ccap%20A_3%29%29)
and from there we find
![P(A_1\cap A_2\cap{A_3}^c)=P((A_1\cap A_2\cap A_3)^c)-P((A_1\cap A_2)^c)=\boxed{0.05}](https://tex.z-dn.net/?f=P%28A_1%5Ccap%20A_2%5Ccap%7BA_3%7D%5Ec%29%3DP%28%28A_1%5Ccap%20A_2%5Ccap%20A_3%29%5Ec%29-P%28%28A_1%5Ccap%20A_2%29%5Ec%29%3D%5Cboxed%7B0.05%7D)
d. The event of having at most two of these defects is complementary to the event that all three defects occur simultaneously:
![P((A_1\cap A_2\cap A_3)^c)=1-P(A_1\cap A_2\cap A_3)=\boxed{0.98}](https://tex.z-dn.net/?f=P%28%28A_1%5Ccap%20A_2%5Ccap%20A_3%29%5Ec%29%3D1-P%28A_1%5Ccap%20A_2%5Ccap%20A_3%29%3D%5Cboxed%7B0.98%7D)