Okay so basically, i’m pretty sure you have to make them catch up with each other, if i’m correct), so if it’s 30 to each student plus 2500 then 2 students is 60 bucks, 3 students is 90 and so on so forth......then if 1 student - 60 then 2 is 120 and vice versa ya know. So keep adding on the amount due for each student plus the extra amount which doesn’t change and make them catch up with each other. Keep on adding student after student and adding the fee until they match up. Sorry it’s not the exact answer but i hope this explains it... if i’m right
Answer:
5 1/4 cups of flour will be in each container
Step-by-step explanation:
21 ÷ 4= 5.25
5.25= 5 1/4
hope this helps babe <3
Answer:
See Below
Step-by-step explanation:
The surface area of cylinder is given by the formula:

Where
r is radius ( diameter is 4, so radius is 4/2 = 2)
h is height ( h = 9)
Lets find original surface are:

<u>Halving diameter:</u>
diameter would be 4/2 = 2, so radius would be 2/2 = 1
So, SA would be:

<u>Halving height:</u>
Height is 9, halving would make it 9/2 = 4.5
Now, calculating new SA:

Original SA is
,
Halving diameter makes it 
Halving height makes it 
So, halving diameter does not have same effect as halving height.
Answer:
<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>
By De morgan's law

which is Bonferroni’s inequality
<h3>Result 1: P (Ac) = 1 − P(A)</h3>
Proof
If S is universal set then

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>
Proof:
If S is a universal set then:

Which show A∪B can be expressed as union of two disjoint sets.
If A and (B∩Ac) are two disjoint sets then
B can be expressed as:

If B is intersection of two disjoint sets then

Then (1) becomes

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>
Proof:
If A and B are two disjoint sets then

<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>
Proof:
If B is subset of A then all elements of B lie in A so A ∩ B =B
where A and A ∩ Bc are disjoint.

From axiom P(E)≥0

Therefore,
P(A)≥P(B)