Selena is 6 years younger than Paul. In 15 years, Paul will be 4 times the age that Selena is now. Find their present ages.
2 answers:
Their present ages :
Paul = 13
Selena = 7
<h3>Further Explanation </h3>One variable linear equation is an equation that has a variable and the exponent number is one.
Can be stated in the form:
or
<h3>ax + b = c, </h3>where a, b, and c are constants, x is a variable
Whereas the two-variable linear equation is a linear equation that has 2 variables and the exponent is one
Can be stated in the form:
x, y = variable
In the statement of the task:
- Selena is 6 years younger than Paul.
can be stated in the equation:
Selena (S) = Paul (P) -6 ⇒S = P-6 ⇒equation 1
- In 15 years, Paul will be 4 times the age that Selena is now.
The equation: 15 + P = 4S⇒equation 2
Equation 1 is substituted into equation 2 so that it becomes:
15 + P = 4 (P-6)
So the solution is:
15+P = 4P-24
15+24 = 3P
39 = 3P
P = 13
From equation 1
S = P - 6
S = 13 - 6
S = 7
<h3>Learn more</h3>
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Keywords : present age, Selena, Paul
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Answer:
The probability is 0.0052
Step-by-step explanation:
Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:
P(A/B) = P(A∩B)/P(B)
The probability P(B) that at least three are aces is the sum of the following probabilities:
- The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
- There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:
So, the number of ways to select exactly 3 aces is:
Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725
Then, the probability P(B) that at least three are aces is:
On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:
P(A∩B) = 1/270,725
Finally, the probability P(A/B) that all four are aces given that at least three are aces is: