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: 14 minutes
Step-by-step explanation:
in the attachment
Answer:
November 13
Step-by-step explanation:
Following dates are given
On November 10 = Merchandise ordered
Date of an invoice prepared, dated and mailed = November 13
Date when the merchandised received by the buyer = November 18
So, the credit period begins when the invoice is prepared, dated and the mailed by the seller to the buyer as it is the evidence of that the merchandise is ordered
150*30% = $45
$150-$45 = $105
Marco still owes his brother $105
Answer:
i think 2x
Step-by-step explanation:
