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:
1.277063499260000000006476094388e+40
Step-by-step explanation:
I used a calculator :)
Answer:
c) 6x - 5y = 15
Step-by-step explanation:
Slope-intercept form of a linear equation:
(where m is the slope and b is the y-intercept)
Maria's line:
Therefore, the slope of Maria's line is
If two lines are perpendicular to each other, the product of their slopes will be -1.
Therefore, the slope of Nate's line (m) is:
Therefore, the linear equation of Nate's line is:
Rearranging this to standard form:
Therefore, <u>option c</u> could be an equation for Nate's line.