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
Answer:
D
Step-by-step explanation:
1, 17, and 20 are very different from the average numbers in this list.
7)
Read my note at the end of problem 5 in another post.
You already know this table represents an exponential function
since each y-coordinate is always the previous y-coordinate multiplied by 6.
That means b = 6, and you have
y = a(6)^x
Now we find "a". When x = 0, y = 5. That means a = 5.
The equation is
y = 5(6)^x
X% of $28,600 is $33,500
x/100 * 28,600 = 33,500
x = 33,500 * 100 / 28,600
x = 117.13% = 117%
117% - 100% = 17%
The answer would be 17%.
Glad to help :)
3 lemons
if you have to show your work 4 divided by 12 is 3
