contestada

According to an automobile association of america report, 9.6% of americans traveled by car over the 2011 memorial day weekend and 88.09% stayed home. what is the probability that a randomly selected american stayed home or traveled by car over the 2011 memorial day weekend? p(stayed home or travelled by car)=

Respuesta :

p(traveled by car)=9.6%
p(stayed home)=88.09%

p(stayed home or travelled by car)=p(stayed home)+p(traveled by car)-p(stayed home and traveled by car)

p(stayed home and traveled by car)=0%

p(stayed home or travelled by car)=88.09%+9.6%-0%
p(stayed home or travelled by car)=97.69%
jdurst04

Answer:

First Part: 0.9769

Second Part: Some Americans may have traveled by other means

Step-by-step explanation:

First Part: You add 9.6 to 88.09 and put it in decimal form.

Second Part: Some Americans may have traveled by other means.

Otras preguntas