Based on the percentage that passed English and those who passed Mathematics and those who failed and passed both, the total number of students who appeared in the examination are 60 students.
The number of students who passed only in Math are 12 students.
<h3>What number of students sat in the exam?</h3>
This can be found as:
= Total who passed English only + Total who passed Math only + Total who failed both + Total who passed both
Assuming the total is n, the equation becomes:
n = 0.75n - 21 + 0.55n - 21 + 21 + 0.05n
n = 1.35n - 21
21 = 0.35n
n = 21 / 0.35
= 60 students
The number who passed mathematics only is:
= (60 x 55%) - students who passed both
= 33 - 21
= 12 students
Find out more on Venn diagrams at brainly.com/question/24581814
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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.
Answer:
Odd
Step-by-step explanation:
One way I remember is that:
Even functions: Have symmetry over the y-axis
Odd functions: Don't have symmetry over the y-axis
The equation given in the question has one unknown variable in the ofrm of "x" and there is also a single equation. So it can be definitely pointed out that the exact value of the unknown variable "x" can be easily determined. Now let us focus on the equation given in the question.
x/35 = 7
x = 35 * 7
x = 245
So we can find from the above deduction that the value of the unknown variable "x" is 245. The correct option among all the options given in the question is option "B". I hope the procedure is not complicated for you to clearly understand.
Answer:
The two graphs are perpendicular
Step-by-step explanation:
The slope of the first one is the reciprocal of the second. Also I graphed it.
