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22. Okay. What we need to do for this one is subtract fraction.l, because we are looking for how much more the second container holds than the first. We need to covert them to like denominators, which in this case is 8 5 3/4 = 5 6/8. Because 6/8 is less than 7/8, we need to regroup by adding 8 and turning the 5 into a 4. The problem would become like this: 4 14/8 - 1 7/8. When you subtract both numbers, you get 3 7/8. There. The second container holds 3 7/8 more gallons of water than the first.
23.
a. 15 student speeches will last 1 1/2 minutes each. In this case, we just simply multiply the fractions by multiply straight across. Multiply the top numbers together and the bottom numbers together. 1 1/2 is 3/2 a an improper fraction. Make 15 have a denominator of 1. 15/1 * 3/2 is 45/2 or 22 1/2 as a mixed number. It will take 22 1/2 minutes to give the speeches.
b. Okay. The teacher makes a 15 minute intro. Let’s add the time it takes to make the speeches and the teacher’s intro. 22 1/2 + 15 0/2 is 37 1/2. Yes. There is enough time to record everyone.
c. There are 60 minutes in 1 hour. Subtract the total amount taken on the camera from 60. 0/2 is less than 1/2. Again, regroup by adding 2 to make 2/2, and crossing out 60, so it becomes 59. 59 2/2 - 37 1/2 is 22 1/2. There are 22 1/2 minutes left on the digital camera.
Answer:
1716 ;
700 ;
1715 ;
658 ;
1254 ;
792
Step-by-step explanation:
Given that :
Number of members (n) = 13
a. How many ways can a group of seven be chosen to work on a project?
13C7:
Recall :
nCr = n! ÷ (n-r)! r!
13C7 = 13! ÷ (13 - 7)!7!
= 13! ÷ 6! 7!
(13*12*11*10*9*8*7!) ÷ 7! (6*5*4*3*2*1)
1235520 / 720
= 1716
b. Suppose seven team members are women and six are men.
Men = 6 ; women = 7
(i) How many groups of seven can be chosen that contain four women and three men?
(7C4) * (6C3)
Using calculator :
7C4 = 35
6C3 = 20
(35 * 20) = 700
(ii) How many groups of seven can be chosen that contain at least one man?
13C7 - 7C7
7C7 = only women
13C7 = 1716
7C7 = 1
1716 - 1 = 1715
(iii) How many groups of seven can be chosen that contain at most three women?
(6C4 * 7C3) + (6C5 * 7C2) + (6C6 * 7C1)
Using calculator :
(15 * 35) + (6 * 21) + (1 * 7)
525 + 126 + 7
= 658
c. Suppose two team members refuse to work together on projects. How many groups of seven can be chosen to work on a project?
(First in second out) + (second in first out) + (both out)
13 - 2 = 11
11C6 + 11C6 + 11C7
Using calculator :
462 + 462 + 330
= 1254
d. Suppose two team members insist on either working together or not at all on projects. How many groups of seven can be chosen to work on a project?
Number of ways with both in the group = 11C5
Number of ways with both out of the group = 11C7
11C5 + 11C7
462 + 330
= 792
Step-by-step explanation:
16 arrangement can be done
