The rate of change between 1990 and 1999 is 15.22
Step-by-step explanation:
The rate of change in this case is given by change in quantity divided by change in time
So,

Hence,
The rate of change between 1990 and 1999 is 15.22
Keywords: Rate of change, median
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Cause a number cube is wack and cant solve nothin bout bowling you gotta use the numbers 6 outta 10
Answer:
B
Step-by-step explanation:
Answer:
2
If R is a relation that is transitive and symmetric, then R is reflexive on dom(R)={a∣(∃b)aRb}: if a∈dom(R), then there is b such that aRb, thus bRa by symmetry, so aRa by transitivity.
Note that if R is symmetric, then dom(R)=range(R)={b∣(∃a)aRb}.
Hence, to get an example of a relation R on a set A that is transitive and symmetric but not reflexive (on A), there has to be some a∈A which is not R-related to any b∈A. There are many examples of this:
A={0,1} and R={(0,0)},
not reflexive on A because (1,1)∉R,
A={0,1,2} and R={(0,0),(0,1),(1,0),(1,1)},
not reflexive on A because (2,2)∉R.
Step-by-step explanation:
Answer:
The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Let 'S' be the sample space associated with the drawing of a card
n (S) = 52C₁ = 52
Let E₁ be the event of the card drawn being a king

Let E₂ be the event of the card drawn being a queen

But E₁ and E₂ are mutually exclusive events
since E₁ U E₂ is the event of drawing a king or a queen
<u><em>step(ii):-</em></u>
The probability of drawing of a king or a queen from a standard deck of playing cards
P( E₁ U E₂ ) = P(E₁) +P(E₂)

P( E₁ U E₂ ) = 
<u><em>step(iii):-</em></u>
The probability of drawing the compliment of a king or a queen from a standard deck of playing cards



<u><em>Conclusion</em></u>:-
The probability of drawing the compliment of a king or a queen from a standard deck of playing cards = 0.846