Find lcm of 28 and 36 which is 252. They will meet again in another 252 days.
ANSWER: $65.63
$50 divided by 4 hours = $12.50 per hour
1/4 can also be written as the fraction 0.25
So 5 1/4 hours is the same as 5.25 hours
$12.50 per hour x 5.25 hours = $65.625 rounded to the nearest cent = $65.63
Answer:


We know that within two deviations from the mean we have 95% of the data from the empirical rule so then below 2 deviation from the mean we have (100-95)/2 % =2.5%. And within 3 deviations from the mean we have 99.7% of the data so then below 3 deviations from the mean we have (100-99.7)/2% =0.15%
And then the final answer for this case would be:

Step-by-step explanation:
For this case we have the following parameters from the variable number of motnhs in service for the fleet of cars

For this case we want to find the percentage of values between :

And we can use the z score formula given by:

In order to calculate how many deviation we are within from the mean. Using this formula for the limits we got:


We know that within two deviations from the mean we have 95% of the data from the empirical rule so then below 2 deviation from the mean we have (100-95)/2 % =2.5%. And within 3 deviations from the mean we have 99.7% of the data so then below 3 deviations from the mean we have (100-99.7)/2% =0.15%
And then the final answer for this case would be:

The first thing we need to do is take in all this information. We already know that the answer is between 0 and 7 because Line Segment AC is equal to 7. Second of all, it is recommended to solve this problem on paper to make it easier. You write the measurement under each individual Line Segment. Now we have to try to find a way how to subtract the Line Segments in order to get the measurement of Line Segment BC. If we were to write an equation. We would get mAB + mBC + mCD + mDE = mAE . We now try to think about the values that they give us to try to simplify it as much as possible. That's all that I can give you. I am sorry. This is a pretty tough and evil question.