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%
The answerrRRRRRRRRRRRRRRRRRRRR is A
Answer:
125
Step-by-step explanation:
5/4=x/100
x=500/4
x=125
Answer: -1
Step-by-step explanation: f(x) is another way of saying y. So when y is 3, according to the table, x is -1
<u>Answer:</u>
- Carly's expenditure was $240 and the number of boxes of cards was required is 24 boxes.
<u>Step-by-step explanation:</u>
We know that:
- 48 + 8(b) = Carly's expenditure
- 10(b) = Carly's expenditure
- 48 + 8b = 10b
- B = Boxes of cards
<u>Work:</u>
- 48 + 8b = 10b
- => 48 + 8b - 8b = 10b - 8b
- => 48 = 2b
- => b = 24
<u>Now, let's substitute the value of b into any equation.</u>
<u>Hence, Carly's expenditure was $240 and the number of boxes of cards was required is 24 boxes.</u>
Hoped this helped.
