Answer:
No invariant point
Step-by-step explanation:
Hello!
When we translate a form, in this case a polygon We must observe the direction of the vector. Since our vector is:
1) Let's apply that translation to this polygon, a square. Check it below:
2) The invariant points are the points that didn't change after the transformation, simply put the points that haven't changed.
Examining the graph, we can see that no, there is not an invariant point, after the translation. There is no common point that belongs to OABC and O'A'B'C' simultaneously. All points moved.
Answer:
a) 0.70
b) 0.82
Step-by-step explanation:
a)
Let M be the event that student get merit scholarship and A be the event that student get athletic scholarship.
P(M)=0.3
P(A)=0.6
P(M∩A)=0.08
P(not getting merit scholarships)=P(M')=?
P(not getting merit scholarships)=1-P(M)
P(not getting merit scholarships)=1-0.3
P(not getting merit scholarships)=0.7
The probability that student not get the merit scholarship is 70%.
b)
P(getting at least one of two scholarships)=P(M or A)=P(M∪A)
P(getting at least one of two scholarships)=P(M)+P(A)-P(M∩A)
P(getting at least one of two scholarships)=0.3+0.6-0.08
P(getting at least one of two scholarships)=0.9-0.08
P(getting at least one of two scholarships)=0.82
The probability that student gets at least one of two scholarships is 82%.
The system shown at the right has no solution, as the grahs never intersect.
On the other hand, the line and the parab. at the left do intersect, and the points of intersection are (-3,0) and (6,6).
