Answer:
1 1/5
Step-by-step explanation:
I have showed the workings and answers in the picture
Answer:
3 of the fish are blue
Step-by-step explanation:
5 fish - 2 fish = 3 fish
<em>hope this helps :)</em>
Answer: 45
<u>Step-by-step explanation:</u>
5 : 5 : 3 is simplified so it can be rewritten as 5x : 5x : 3x
↓ ↓ ↓
Sara Nick June
Sara has 18 more than June --> 5x = 3x + 18
2x = 18
x = 9
In expanded form we have 5(9) : 5(9) : 3(9)
45 : 45 : 27
↓ ↓ ↓
Sara Nick June
Answer:
D) 
.
Step-by-step explanation:
Given : If the first term of the series is 30 and the 14th term is 95,
To find : what is the sum of all the terms of the series.
Solution : We have given
First term = 30 .
14 th term = 95.
Sum of all term = 
.
Here, n = 14.
.
.
.
.
Therefore, D) .
Answer:
a) 1/2
b) 1/n
c) 1/4
Step-by-step explanation:
a) For each permutation, either 1 precedes 2 or 2 precedes 1. For each permutation in which 1 precedes 2, we can swap 1 and 2 to obtain a permutation in which 2 preceds 1. Thus, half of the total permutations will involve in 1 preceding 2, hence, the probability for a permutation having 1 before 2 is 1/2.
c) If 2 is at the start of the permutation, then it is impossible for 1 to be before 2. If that is not the case, then 1 has a probability of 1/n-1 to be exactly in the position before 2. We can divide in 2 cases using the theorem of total probability,
P( 1 immediately preceds 2) = P (1 immediately precedes 2 | 2 is at position 1) * P(2 is at position 1) + P(1 immediately precedes 2 | 2 is not at position 1) * P(2 is not at position 1) = 0 * 1/n + (1/n-1)*(n-1/n) = 1/n.
d) We can divide the total of permutations in 4 different groups with equal cardinality:
- Those in which n precedes 1 and n-1 precedes 2
- those in which n precedes 1 and 2 precedes n-1
- those in which 1 precedes n and n-1 precedes 2
- those in which 1 precedes n and 2 precedes n-1
All this groups have equal cardinality because we can obtain any element from one group from another by making a permutations between 1 and n and/or 2 and n-1.
This means that the total amount of favourable cases (elements of the first group) are a quarter of the total, hence, the probability of the event is 1/4.
