probability of one go on vacation and one does not go on vacation is 0.2356 and probability none of them go on vacation is 0.1444
According to a survey, 62% of americans go on vacation each .
two americans are chosen from a group of 100 americans.
what is the probability that one or both of the people chosen does not go on vacation each ?
62% americans go on vacation t
hen 38% americans does not go on vacation
Total group of americans is 100
out of 100 americans 62 go on vacation and 38 does not go on vacation
probability of one go on vacation and one does not go on vacation
= 0.62*0.38
=0.2356
probability none of them go on vacation
= 0.38*0.38
=0.1444
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Answer:
Tony make 4 half-court shots and misses 16 half-court shots.
Step-by-step explanation:
Given:
Points earned for every successful, half-court shots = 
Points deducted for missing the shots = 
or 
Total number of shots in the game =
Final points earned by Tony = 
According to the question:
Let the number of successful half-court shot be 
And the number of shots missed be 
So,
In terms of points we can re-write it as:
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Tony make 'x' shots that is 4 and the number of shot he missed is (20-x) =(20-4)=16
I believe the answers to be 75+5x=y
Mary = (1/3)x
Erin = x - 8
John = x
Here is your equation:
(1/3)x + (x - 8) + x = 267
Take it from here.
