Answer: 10:55
Step-by-step explanation:
Taking statement at face value and the simplest scenario that commencing from 08:00am the buses take a route from depot that returns bus A to depot at 25min intervals while Bus B returns at 35min intervals.
The time the buses will be back at the depot simultaneously will be when:
N(a) * 25mins = N(b) * 35mins
Therefore, when N(b) * 35 is divisible by 25 where N(a) and N(b) are integers.
Multiples of 25 (Bus A) = 25, 50, 75, 100, 125, 150, 175, 200 etc
Multiples of 35 (Bus B) = 35, 70, 105, 140, 175, 210, 245 etc
This shows that after 7 circuits by BUS A and 5 circuits by Bus B, there will be an equal number which is 175 minutes.
So both buses are next at Depot together after 175minutes (2hr 55min) on the clock that is
at 08:00 + 2:55 = 10:55
-1.6 - 0.3
I hope this hepled you and that the answer is correct
✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧
Hello!
✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧
❖ 8.5 in = 425 miles
If 1 mi = 50 inches, multiply by 8.5:
50 x 8.5 = 425
~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡
~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ
Hi, Deedee. I am quite confused with the numbers that you've given because they don't add up to 100. Nevertheless, if by 'this', you mean the Thanksgiving culture, then you would just count the English and the Native American heritage. This is because the first Thanksgiving dinner was shared between the English colonists and the Native American tribes.
I hope I was able to help you in a way. Have a good day.
Answer:
The 90% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.
Step-by-step explanation:
We have the standard deviation for the sample, so we use the t-distribution to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 9 - 1 = 8
90% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 8 degrees of freedom(y-axis) and a confidence level of 
. So we have T = 1.8595
The margin of error is:
M = T*s = 1.8595*2.7 = 5
In which s is the standard deviation of the sample.
The lower end of the interval is the sample mean subtracted by M. So it is 25.3 - 5 = 20.3 milligrams
The upper end of the interval is the sample mean added to M. So it is 25.3 + 5 = 30.3 milligrams.
The 90% confidence interval for the mean nicotine content of this brand of cigarette is between 20.3 milligrams and 30.3 milligrams.
