Answer: 15 months
Step-by-step explanation:
Her balance after paying the registration fee is $300. Then we need to divide this remaining money into as many chunks (months )of $20. So we do 300/20 = 15 months
Answer:
They traveling at
in backward direction.
Step-by-step explanation:
Given mass of Ammy 
and velocity is forward at 
Also, mass of Jenny 
Who is travelling in opposite direction 
Let velocity in forward direction is positive and velocity in backward direction is negative.
Also, velocity after crash is 
As they crashed into each other and they moved as one mass. We can apply principle of conservation of momentum.
Conservation of momentum says momentum before collision should equal to momentum after collision.


So, velocity after collision
≅
is in opposite direction. That is the direction of Jenny.
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)
Answer:
B) The heights of the bear should equal the class frequency.
Step-by-step explanation:
In drawing a histogram, the heights of the bear should equal the class frequency. as in histogram height of bar represent the frequency density and it´s area represent the frequency of class interval. Frequency distribution are represented by mean of rectangle. Earlier in the bar graph, width of bar does not represent any information, however, histogram´s bar width represent class interval.