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:
product A on machine 1 would make 27 units (rounded down) 81 dollars profit.
product A on machine 2 would make 25 units (rounded down) 75 dollars profit.
product b on machine 1 would make 15 units (rounded down) 75 dollars profit.
product b on machine 2 would make 16 units (rounded down) 80 dollars profit.
Step-by-step explanation:
a1 refers to product a on machine 1
a2 refers to product a on machine 2
same rule applies for product b 1 and 2
Answer:
11.691
Step-by-step explanation:
Too lazy to wright, sorry.
Answer and explanation:
Benchmark fractions are fractions that are used as references in measuring other fractions. They are easily estimated and so can be used in measuring more "specific" fractions such as 1/5, 7/9, 3/7, 1/3 etc. If I wanted to measure 1 1/3cm for instance using a calibrated ruler, having centimeter measurements, I would first find 1cm on the ruler and then find half of one centimeter. Seeing that half is bigger than 1/3 but close, I could then estimate 1/3 to be somewhere less than 1/2 but a bit close to it
Answer:
2(Jill) - 20 + (Jill) = 205
3(Jill) = 225
(Jill) = 75
(Jack) = 2(75) - 20 => 150 - 20 = 130
Jill= 75
Jack= 130
Step-by-step explanation:
