How in the heck do I do this?
1 answer:
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So... the radiator has 15 liters of 70% antifreeze.. but needs an 80% antifreeze
well, so, you need to drain some and put some with higher percentage, seems to be, you will end up at the same 15 liters, possible the radiator's capacity, of 80% antifreeze
so, the same amount going out, of 70% is the same amount going in, of 100% antifreeze
now.. let's use the decimal format for the percents, or 70% is 70/100 or 0.7 and so on
so.. let's subtract, from the current solution, 0.7x and add 1x or x, our antifreeze concentration amount, should be 12 though
10.5 - 0.7x + x = 12
solve for "x"
well, so, you need to drain some and put some with higher percentage, seems to be, you will end up at the same 15 liters, possible the radiator's capacity, of 80% antifreeze
so, the same amount going out, of 70% is the same amount going in, of 100% antifreeze
now.. let's use the decimal format for the percents, or 70% is 70/100 or 0.7 and so on
so.. let's subtract, from the current solution, 0.7x and add 1x or x, our antifreeze concentration amount, should be 12 though
10.5 - 0.7x + x = 12
solve for "x"
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