25.8%
First, determine how many standard deviations from the norm that 3 tons are. So:
(3 - 2.43) / 0.88 = 0.57/0.88 = 0.647727273
So 3 tons would be 0.647727273 deviations from the norm. Now using a standard normal table, lookup the value 0.65 (the table I'm using has z-values to only 2 decimal places, so I rounded the z-value I got from 0.647727273 to 0.65). The value I got is 0.24215. Now this value is the probability of getting a value between the mean and the z-score. What I want is the probability of getting that z-score and anything higher. So subtract the value from 0.5, so 0.5 - 0.24215 = 0.25785 = 25.785%
So the probability that more than 3 tons will be dumped in a week is 25.8%
Answer:
8x-24
Step-by-step explanation:
8(x-3)
8 times x= 8x
8 times-3= -24
Yes because fraction wise the denominator that is the smallest is greater. At least that is what I learned
Determining a car's depreciation over a ten year period is considered a bivariate.
<h3>What is a bivariate?</h3>
A Bivariate data is made up of two variables that are observed against each other. In determining the deprecation of a car, the cost of the car is observed against the passage of time and the depreciation factor.
To learn more about depreciation, please check: brainly.com/question/25552427
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