Asked by Anonymous
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?
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?
Answers
Answered by
bobpursley
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
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
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