Cameron is older than Hassan. Their ages are consecutive integers. Find Cameron's
1 answer:
Answer:
25
Step-by-step explanation:
So consecutive integers, just means they're separated by a value of 1. This can be generally expressed as "a, a+1" where these two values would be consecutive integers assuming "a" is an integer.
So let's express Hassan's age as the variable "x", since it's unknown. Since Cameron is older, and by definition of a consecutive integer, Cameron's can be expressed as "x+1"
So the equation we need to set up is Cameron's age + 5(Hassan's age) = 145
So we can substitute the variables we defined to express Cameron and Hassan's age:
Distribute the 5:
Add like terms:
Subtract 1 from both sides:
Divide both sides by 6:
Since we used "x" to represent Hassan's age, Hassan's age is 24. Since we used "x+1" to represent Cameron's age, Cameron's age is "24+1" which is just 25
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Answer:
the probability the die chosen was green is 0.9
Step-by-step explanation:
Given that:
A bag contains two six-sided dice: one red, one green.
The red die has faces numbered 1, 2, 3, 4, 5, and 6.
The green die has faces numbered 1, 2, 3, 4, 4, and 4.
From above, the probability of obtaining 4 in a single throw of a fair die is:
P (4 | red dice) =
P (4 | green dice) = =
A die is selected at random and rolled four times.
As the die is selected randomly; the probability of the first die must be equal to the probability of the second die =
The probability of two 1's and two 4's in the first dice can be calculated as:
=
=
=
=
=
=
The probability of two 1's and two 4's in the second dice can be calculated as:
=
=
=
=
=
∴
The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )
The probability of two 1's and two 4's in both die =
The probability of two 1's and two 4's in both die =
The probability of two 1's and two 4's in both die =
By applying Bayes Theorem; the probability that the die was green can be calculated as:
P(second die (green) | two 1's and two 4's ) = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)
P(second die (green) | two 1's and two 4's ) =
P(second die (green) | two 1's and two 4's ) =
P(second die (green) | two 1's and two 4's ) = 0.9
Thus; the probability the die chosen was green is 0.9