Find the value of x so that the ratios 8: x and 12: 18 are equivalent.
2 answers:
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Answer:
x cannot be 5 or -3.
Step-by-step explanation:
The restrictions for a fraction is that the bottom cannot be 0.
So if we find when the bottom is 0 we have found the values that x cannot be.
Let's solve x^2-2x-15=0.
Since the coefficient of x^2 is 1 all we have to do is find two numbers whose product is -15 and whose sum is -2.
Those numbers are -5 and 3 since (-5)(3)=-15 and (-5)+(3)=-2.
So the factored form of the equation is:
(x-5)(x+3)=0
This means either x-5=0 or x+3=0.
We do have to solve both.
x-5=0 can be solved by adding 5 on both sides.
x-5+5=0+5
x+0=0+5
x=5
x+3=0 can be solved by subtracting 3 on both sides.
x+3-3=0-3
x+0=0-3
x=-3
So x can be any number except x=5 or x=-3.
We already factored the bottom as (x-5)(x+3).
The top is a difference of squares, x^2-a^2,
which can be factored as (x-a)(x+a).
So the top factors as (x-5)(x+5).
The fraction can then be written as:
This can further simplified assuming x is not 5 we can write it as . I canceled the common factor of (x-5).
Answer:
For the pairs that satisfy that the sum is less than 7 we have:
(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)
So we have a total of 15 pairs where the sum is less than 7
And for an odd sum we have the following pairs
(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)
A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got:
Step-by-step explanation:
For this case when a pair of dice is rolled we have the following sample pace for the outcomes:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
We see 36 possible outcomes and we want to find how many of these pairs we got rolling an odd sum or a sum less than 7
For the pairs that satisfy that the sum is less than 7 we have:
(1,1) (2,1) (1,2) (3,1) (2,2) (1,3) (4,1) (3,2) (2,3) (1,4) (5,1) (4.2) (3,3) (2,4) (1,5)
So we have a total of 15 pairs where the sum is less than 7
And for an odd sum we have the following pairs
(2,1) (1,2) (4,1) (3,2) (2,3) (1,4) (6,2) (5,2) (4,3) (3,4) (2,5) (1,6) (6,3) (5,4) (4,5) (3,6) (6,5) (5,6)
A total of 18 pairs and we have 6 pairs who are in both cases for the sum less than 7 and the sum an odd number that represent the intersection and using the total probability rule we got: