Answer:
Danny will watch all of them on the same day again on August 31
Step-by-step explanation:
Here, we have a word problem.
The shows and frequency of watching are;
Show A Every 2 days
Show B Every 3rd day
Show C Every 5th day
Show D Every 10th day
Danny watched all four shows on August 1, therefore, the day he will watch all four shows can be found by finding the Lowest Common Multiple, LCM of the frequency of his watching each of the shows. That is, the LCM of 2, 3, 5 and 10
Therefore we have, since 10 is the highest frequency and 3 is not a factor of 10, then the LCM that applies to 3 and 10 is 3×10 = 30.
Since the other frequencies are also a factor of 30, then the LCM of 2, 3, 5, and 10 is 30
Therefore, Danny will watch all four shows again on the same day again in 30 days after August 1. That is August (1 + 30) = August 31.
Answer:
they both have 5 in common
if you were to reduce the fraction it would be 11/16 because
55 divided by 5 is 11 and 80 divided by 5 is 16
Step-by-step explanation:
Answer:
At the rate expected, 125 of the 10,000 cars from said manufacturer will contain a defect.
Step-by-step explanation:
Given a sample size of 400, this will be a representation of the total of all cars from this manufacturer.
If 5 of these cars have been found to have a defect, then the rate of defects within the company's cars is expected to be
5/400
Using this proportion to expand to a sample size of 10,000 cars then, we can create the expression
5/400 = x/10,000 Multiply both sides by 10,000 to isolate x
50,000/400 = x Reduce the fraction
125 = x
Therefore, within a sample of 10,000 cars from said manufacturer, it can be expected that 125 of these vehicles will have a defect.
