Answer:
<em> n = 13 </em>
Step-by-step explanation:
=
+ (n - 1)d
= 3 + 7(n - 1) (for the first AP)
= 63 + 2(n - 1) ( for the second one)
3 + 7(n - 1) = 63 + 2(n - 1)
3 + 7n - 7 = 63 + 2n - 2
5n = 65
<em>n = 13</em>
4 - x +
-formula"> 
21
1 answer:
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Answer:
Step-by-step explanation:
Given the numbers on the chips = 1, 1, 3 and 5
Miguel chooses two chips.
Condition of winning: Both the chips are same i.e. 1 and 1 are chosen.
Miguel gets $2 on winning and loses $1 on getting different numbers.
To find:
Probability of winning $2 and losing $1 respectively.
Solution:
Here, we are given 4 numbers 1, 1, 3 and 5 out of which 2 numbers are to be chosen.
This is a simple selection problem.
The total number of ways of selecting r numbers from n is given as:
Here, n = 4 and r = 2.
So, total number of ways =
Total number of favorable cases in winning = choosing two 1's from two 1's i.e.
Now, let us have a look at the formula of probability of an event E:
So, the probability of winning.
Total number of favorable cases for -1: (6-1) = 5
So, probability of getting -1:
Please refer to the attached image for answer table.
