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: 97
Step-by-step explanation:
So it is a 1/26 chance of drawing c first. Then its the same chance for A and same chance for R. You multiply these fractions so the answer is 1/17576. If this is right could you possibly give me brainliest? Hope this helped.
36 is the number that makes it completely factored
Answer:
(a)23 (b)90 (c)3
Step-by-step explanation:
The equation for the line of best fit for this situation is given as
where x=average temperature in degrees
y=average number of hot dogs she sold,
(a) The expected number of hot dogs sold when the temperature is 50° would be___hot dogs.
When x=50°
When the temperature is 50°, the expected number of hot dogs sold would be 23.
(b)If the vendor sold 35 hot dogs, the temperature is expected to be ___degrees.
If y=35
Multiply both sides by 10/3
If the vendor sold 35 hot dogs, the temperature is expected to be 90 degrees.
(c) Based on the line of best fit, for every 10-degree increase in temperature, she should sell 3 more hot dogs.