You score an 84 on the first of two exams. Write and solve an equation to find the score x that you need on the second exam to h
1 answer:
You might be interested in
Answer:
(a) Hypergeometric distribution
(b) 0.785384...
Step-by-step explanation:
(a) looking at the question, we see that , suppose Evans studied s key terms, this implies that he has not studied 100 - s key terms. Suppose
if X = k, then k out of the s key terms he had studied appeared. But 10 - k out of 100 - s key terms he hasn't studied appeared. Thus X is an Hypergeometric distribution, X~HGeom(s, 100 - s, 10) with PMF:
Answer:
Option A
x=6 , y=9
Step-by-step explanation:
As we can see in the figure that point T is the mid point of rectangle and there are two diagonals QS and PR
So in basic geometry we know that the length of the diagonal from the mid point i.e. Taking the diagonal QS the length of QT and TS would be equal because they are same diagonal which is cut into two parts by point T
Similarly the Length of RT and the length of TP would be same
From all this we deduct that
3x = 2y ..................(i)
and
y + 3 = 2x ................(ii)
From equation (i)
3x = 2y
Dividing both sides by 3 the equation becomes
x= ....................(iii)
Putting this value in equation (ii)
y + 3 = 2()
y+3 =
multiplying both sides by 3
3(y+3)=
the equation becomes
3y+9=4y
subtracting 3y from both sides
3y-3y+9=4y-3y
y=9
Putting the value of y in equation (iii)
x=
solving it
x=
x=6
so x=6 and y=9 is answer which is option A