Answer:
The probability that the household has only cell phones and has high-speed Internet is 0.408
Step-by-step explanation:
Let A be the event that represents U.S. households has only cell phones
Let B be the event that represents U.S. households have high-speed Internet.
We are given that 51% of U.S. households has only cell phones
P(A)=0.51
We are given that 70% of the U.S. households have high-speed Internet.
P(B)=0.7
We are given that U.S. households having only cell phones, 80% have high-speed Internet. A U.S household is randomly selected.
P(B|A)=0.8

Hence the probability that the household has only cell phones and has high-speed Internet is 0.408
-4+3x= 2x +3x +3
-4-3= 2x + 3x -3x
-7 =2x
X= -7/2
Answer:
-12 < p
Step-by-step explanation:
28 > 4 – 2p
Subtract 4 from each side
28-4 > 4-4 – 2p
24 > -2p
Divide each side by -2, remembering to flip the inequality
24/-2 < -2p/-2
-12 < p