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: Move terms to the left side−52+3=−9−5x2+3x=−9−52+3−(−9)=0−
Common factor−52+3+9=0−5x2+3x+9=0−(52−3−9)=0
Divide both sides by the same factor−(52−3−9)=0−(5x2−3x−9)=052−3−9=0
Solution=3±321 over 10
Step-by-step explanation:
To reduce a fraction, divide the numerator and the denominator equally until they reach the simplest whole number possible.
In this case, the numerator (720) and the denominator (1080) can both be divided by 360 to get 2/3, our reduced fraction.
Answer:
The IQR is given by:
If we want to find any possible outliers we can use the following formulas for the limits:
And if we find the lower limt we got:
So then the left boundary for this case would be 3 days
Step-by-step explanation:
For this case we have the following 5 number summary from the data of 144 values:
Minimum: 9 days
Q1: 18 days
Median: 21 days
Q3: 28 days
Maximum: 56 days
The IQR is given by:
If we want to find any possible outliers we can use the following formulas for the limits:
And if we find the lower limt we got:
So then the left boundary for this case would be 3 days
Answer:
There is not enough information to answer, you did not say measurements of both drinks.
Step-by-step explanation:
