1. Here is the frequency table for this distribution (the frequency histogram is not shown):
Since I cannot make a table on this forum,the format will be like this: Time(minutes)and Number of telephone calls.
0<t<=5 and 12 telephone calls
5<t<=10 and 4 telephone calls
10<t<=15 and 6 telephone calls
15<t<=20 and 8 telephone calls
A. Write down the MID INTERVAL VALUE (what does this mean?) of the 10<t<=15.
B. Use your graphic display calculator to find an estimate for the mean time.
I got 11.16777 but the answer key says 9.16777...
2. 50 students total.
12 students travelled by car only
7 students travelled by bus only
5 students travelled by car and walked,but did not use a bus
10 students travelled by bus and walked,but did not use a car
3 students used all three forms of travel
I drew a Venn diagram.
Two students are chosen at random from all 50 students.
- Find the probability that
a) Both students walked
b) only one of the students walked
3. What does (dA)/(dr) mean in the surface area of cylinder the problem?
1.
a. On the interval from 10 to 15, the mid interval value for me is 6 per five minutes.
b. I don't know how you got your number. If I were calculating, I would do a mid interval of
2.5*12+7.5*4+ 12.5*6+ 17.5*8 all divided by the total number of calls.
2
a. Pr=walking students/total students
= (3+5+10)/(37) * (17/36)
a. pr=wlked*not walked
= 18/37 * (19/35)
check that
1. A. The MID INTERVAL VALUE of the 10 < t ≤ 15 interval refers to the middle value within that interval. It can be calculated by finding the average of the lower and upper values of the interval. In this case, the lower value is 10 and the upper value is 15. The MID INTERVAL VALUE can be calculated as (10 + 15) / 2 = 12.5.
B. To estimate the mean time using a graphic display calculator, we need to calculate the weighted mean. Multiply each mid interval value (from the frequency table) by its corresponding frequency. Then, sum up these products and divide it by the total number of telephone calls.
Using the given table:
(5 * 12) + (10 * 4) + (15 * 6) + (20 * 8) = 60 + 40 + 90 + 160 = 350
Total number of telephone calls = 12 + 4 + 6 + 8 = 30
The estimate for the mean time will be 350 / 30 = 11.66777, rounding to the nearest decimal place would be 11.7. Therefore, your calculation of 11.16777 is closer to the correct answer than 9.16777.
2. a) To find the probability that both students walked, we need to consider the number of students who walked and divide it by the total number of possible outcomes when two students are chosen randomly from the group of 50.
From the Venn diagram, 5 students traveled by car and walked, but did not use a bus. Therefore, there are 5 students who walked. The probability of the first student walking would be 5/50, and for the second student, after one student has already been chosen, the probability would be 4/49.
To find the probability of both students walking, we multiply these probabilities:
P(both students walked) = (5/50) * (4/49) = 0.01633
Therefore, the probability that both students walked is approximately 0.01633.
b) For only one of the students to walk, we need to consider three possibilities: the first student walks while the second student does not, the second student walks while the first does not, or both students walk, but one also uses another mode of transportation.
First student walks, second doesn't:
P(first student walks) = 5/50
P(second student doesn't walk) = 45/49
P(only one walks, and it's the first student) = (5/50) * (45/49) = 0.04592
Second student walks, first doesn't:
P(first student doesn't walk) = 45/50
P(second student walks) = 5/49
P(only one walks, and it's the second student) = (45/50) * (5/49) = 0.04592
Both students walk, but one uses another mode of transportation:
P(both students walk) = 0.01633 (from part a)
P(one uses another mode of transportation) = 3/50
P(only one walks, and uses another mode of transportation) = (0.01633) * (3/50) = 0.00098
Finally, we add up these three probabilities to find the total probability that only one of the students walked:
P(only one of the students walked) = 0.04592 + 0.04592 + 0.00098 = 0.09282
Therefore, the probability that only one of the students walked is approximately 0.09282.
3. In the context of the surface area of a cylinder problem, (dA)/(dr) refers to the rate of change of the surface area (A) with respect to the radius (r). It indicates how the surface area of the cylinder changes when the radius is altered.
By taking the derivative of the surface area equation with respect to the radius, you can determine the exact mathematical expression for (dA)/(dr). This derivative will provide you with a relationship between the change in surface area and the change in radius.
Knowing (dA)/(dr) will allow you to understand the impact of modifying the radius on the surface area, helping you solve problems related to optimizing or analyzing changes in the surface area of a cylinder.