Answer: The first experiment has M probabilities, and the second has I(m) outcomes, that depends on the result of the first.
And lets call m to the result of the first experiment.
If the outcome of the first experiment is 1, then the second experiment has 1 possible outcome.
If the outcome of the first experiment is 2, then the second experiment has 2 possibles outcomes.
If the outcome of the first experiment is M, then the second experiment has M possibles outcomes.
And so on.
So the total number of combinations C is the sum of all the cases, where we exami
1 outcome for m = 1
+
2 outcomes for m=2
+
.
.
.
+
M outcomes for m = M
C = 1 + 2 + 3 + 4 +...´+M
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
Answer:
The magazine that cost $15.24 for 6 issues cost more by $0.60/60 cents.
Step-by-step explanation:
Since $15.24 is 6 issues each and you need to even out the issues add $15.24+$15.24 then you would get $30.48, because its 15.24 each 6 add them both and it would be 12 issues. And the first issue $29.88 for 12 issues is only $29.88 while the other is $30.48 so the one with 6 issues/the second cost more than the first.
60 cents more because 29.88+60=30.48.
Answer:
10⁵ billion
Step-by-step explanation:
given that distance from sun to earth
= 1.5 x 10⁸ km (we need to convert this to millimeters)
= 1.5 x 10⁸ km x 1 million mm per km
= 1.5 x 10⁸ km x 1,000,000 mm/km
= 1.5 x 10⁸ km x 10⁶ mm/km
= 1.5 x 10⁸ x 10⁶
= 1.5 x 10⁽⁸ ⁺ ⁶⁾
= 1.5 x 10¹⁴ mm
also given that a quarter is 1.5mm thick, hence the number of quarters that will span the distance between the sun an earth (i.e 1.5 x 10¹⁴ mm)
= 1.5 x 10¹⁴ mm ÷ 1.5 mm/quarter
= 10¹⁴
= 100,000,000,000,000 (remember that a billion has 9 zeros)
= 100,000 billion
= 10⁵ billion
Answer: (f + 2) (f + 8)
It's hard to explain, but I suggest you use the AC method. Hope this helps!
