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.
5 divided by 4=1.25 I think I am right
Answer:
The bookshelf will fit between the windows with 1 inch remaining.
Step-by-step explanation:
Total width available for bookshelf
Six and half feet = 6.5 feet = 78 inches
Total width of the bookshelf = 77 inches
Hence, the bookshelf Mrs. Aguilar has purchased will fit between the windows with 1 inch remaining.
Answer: 1.3 is her interest rate hopefully I'm correct on this
Answer:
a. 80 students
b. 92 students
Step-by-step explanation:
Represent arts students with A and Dance students with D.
So, we have,
n(A) = 35
n(D) = 57
Required
Determine n(A or D)
Solving (a):
Here, we have:
n(A and D) = 12
n(A or D) is calculated as thus:
n(A or D) = n(A) + n(D) - n(A and D)
n(A or D) = 35 + 57 - 12
n(A or D) = 80
b. From the given details
n(A and D) = 0 because both students are not mixed up as in (a) above
Using the same formula as (a).
n(A or D) = n(A) + n(D) - n(A and D)
n(A or D) = 35 + 57 - 0
n(A or D) = 92
