solution
this is what I did. total number of ways without any restriction =900. we have to subtract the case when at least one child doesn't get any fruit. let that be C1∪C2∪C3 and we use the principle of mutual inclusion and exclusion to find that. C1∪C2∪C3=C1+C2+C3−(all two intersections)+3 intersections (no one gets a fruit case) =3⋅60−3+1=178 so case where each child gets at least one fruit =900−178=722.
Is this right, someone saying answer =49
We know that
The intersection point of both graphs is a common point for both functions, for which for the same input value, both functions will have the same output value.
so
the point of intersection is
for an input value equal to
the output value for both functions is
therefore
the answer is the option
X= 4
Answer:
6
Step-by-step explanation:
9-3=6
Answer:
A)3
Step-by-step explanation:
