Answer:
1 13/28
Step-by-step explanation:
3×4/28 + 5×4/28. it 28 because the lcm of 4 and 7 is 28
21/28+20/28=41/28=1 13/28
We can set up an equation with x as the price of one cup:
4x=0.96
x=0.96/4=0.24
Each cup is $0.24
Hope this helped!
Answer:
the answer is C
Step-by-step explanation:
Answer: Option C)1 over 15 minus 1 over x equals 1 over 20
Explanation:
Since, Micah can fill a box with books in 15 minutes.
Therefore, the work done by Micah in one minute= 1/15
Also, Sydney takes the books out puts them on a shelf.
And the times taken by Micah when Sydney is also taking the books outside from the self= 20 minutes
Therefore, the work done by Micah in one minute when Sydney taking books out of the box= 1/20
Let Sydney alone takes x minutes to take books outsides the shelf.
Then, work done by Sydney in one minute=1/x
Thus, the work done by Sydney( by taking books out of the box)= the work done by Micah - work done by Micah and Sydney simultaneously= 1/15-1/20
⇒1/x=1/15-1/20
⇒1/15-1/20=1/x
⇒1/15-1/x=1/20 is the required expression.
Therefore, Option C is correct.
Explanation:
Marginal distribution: This distribution gives the probability for each possible value of the Random variable ignoring other random variables. Basically, the values of other variables is not considered in the marginal distribution, they can be any value possible. For example, if you have two variables X and Y, the probability of X being equal to a value, lets say, 4, contemplates every possible scenario where X is equal to 4, independently of the value Y has taken. If you want the probability of a dice being a multiple of 3, you are interested that the dice is either 3 or 6, but you dont care if the dice is even or odd.
Conditional distribution: This distribution contrasts from the previous one in the sense that we are restricting the universe of events to specific condition for other variable, making a modification of our marginal results. If we know that throwing a dice will give us a result higher than 2, then to in order to calculate the probability of the dice being a multiple of 3 using that condition, we have two favourable cases (3 and 6) from 4 total possible results (3,4,5 and 6) discarding the impossible values (1 and 2) from this universe since they dont match the condition given (note that the restrictions given can also reduce the total of favourable cases).
The joint distribution calculates the probabilities for two different events (related to two different random variables) occuring simultaneously. If we want to calculate the joint probability of a dice being multiple of 3 and greater than 2 at the same time, our possible cases in this case are 3 and 6 from 6 possible results. We are not discarding 1 or 2 as possible results because we are not assuming, that the dice is greater than 2, that is another condition that we should met in the combination of events.
