Asa has 15 dimes and 32 quarters.
Step-by-step explanation:
Given,
Total coins = 47
Worth of coins = $9.50 = 9.50*100 = 950 cents
1 dime = 10 cents
1 quarter = 25 cents
Let,
Number of dimes = x
Number of quarters = y
According to given statement;
x+y=47 Eqn 1
10x+25y=950 Eqn 2
Multiplying Eqn 1 by 10

Subtracting Eqn 3 from Eqn 2

Dividing both sides by 15

Putting y=32 in Eqn 1

Asa has 15 dimes and 32 quarters.
Keywords: linear equation, elimination method
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Answer:
(-5, 2)
Step-by-step explanation:
<u>plug x into the equation x - 5y = -15</u>
-5 - 5y = -15
add 5 to both sides: -5y = -10
divide by -5: y = 2
Dy/dx=1/2, 1/2, 1/2 etc
So this is a linear equation of the form y=mx+b where m=dy/dx=1/2 so
y=x/2 +b, now we can use any point to solve for the y-intercept, "b", I'll use (7,0)
0=7/2 +b
b=-7/2 so
y=x/2-7/2
y=(x-7)/2
We're told that



where the last fact is due to the law of total probability:



so that
and
are complementary.
By definition of conditional probability, we have



We make use of the addition rule and complementary probabilities to rewrite this as


![\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)](https://tex.z-dn.net/?f=%5Cimplies%20P%28B%29-%5B1-P%28A%5Ccup%20B%29%5EC%5D%3D%5B1-P%28B%29%5D-P%28A%5Ccup%20B%5EC%29)
![\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]](https://tex.z-dn.net/?f=%5Cimplies2P%28B%29%3D2-%5BP%28A%5Ccup%20B%29%5EC%2BP%28A%5Ccup%20B%5EC%29%5D)
![\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]](https://tex.z-dn.net/?f=%5Cimplies2P%28B%29%3D%5B1-P%28A%5Ccup%20B%29%5EC%5D%2B%5B1-P%28A%5Ccup%20B%5EC%29%5D)


By the law of total probability,


and substituting this into
gives
![2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]](https://tex.z-dn.net/?f=2P%28B%29%3DP%28A%5Ccup%20B%29%2B%5BP%28B%29-P%28A%5Ccap%20B%29%5D)

