An engineer spends 30% of the time at home and 40% at work, 6 mi away. The remaining 30% is spent commuting between home and wor
k in a straight line at a steady velocity. Calculate the probability that at a randomly chosen time the engineer is at least 2 mi from home.
1 answer:
Answer:
1
Step-by-step explanation:
Probability is given by number of possible outcomes ÷ number of total outcomes
The time commuting between home and work is 6mi
Probability (the engineer is at least 2mi from home) means that the engineer is either 2mi from home or more than 2mi from home
Probability (the engineer is 2mi from home) = 2/6
Probability (the engineer is more than 2mi from home) = 1 - 2/6 = (6-2)/6 = 4/6
Probability (the engineer is at least 2mi from home) = 2/6 + 4/6 = 6/6 = 1
You might be interested in
X=0,2,3/8
That is what x equal i don’t know the question but I hope this help
Answer:
huhhh??
Step-by-step explanation:
i didn't unxerstand the question
Step-by-step explanation:
4x + 8y = 40
4x + (8×0.8) = 40
4x + 6.4 = 40
4x = 33.6
4x÷4 = 33.6 ÷ 4
x = 8.4
Answer:
a. $3.36
b. $70.86
Step-by-step explanation:
5% of $67.50 is $3.36
$67.50 + $3.36 = $70.86