Make 2 equations one for the price and one for quantity. so first set x = AA and y= AAA. for price 37= 1x+.75y. now for the quantity we have x+y=42. Now we can solve by subtracting the equation for price from the equation for quantity and we get .25y =5. solving this we get y= 20. Now we plug that value back in and solve for x. 22 double A's and 20 triple A's.
Answer:
The winner completed the race in 96 hours and 52 minutes.
Step-by-step explanation:
Given:
Distance of the cycling race = 2292 miles
Average speed of the winner = 
We need to find time required by winner to complete the race.
Solution:
Now we know that;
Time required can be calculated by dividing Total Distance from the Average speed.
framing in equation form we get;
time required by winner to complete the race = 
Now converting 
into minutes we get;
Hence the winner completed the race in 96 hours and 52 minutes.
Answer:
AC = 2.44
Step-by-step explanation:
Reference angle = 26°
Opposite side = AC = ?
Adjacent side = BC = 5
Applying TOA, we have:
Tan 26 = opp/adj
Tan 26 = AC/5
multiply both sides by 5
Tan 26 × 5 = AC
AC = 2.44 (nearest hundredth)
Answer: Yes it is.
Step-by-step explanation: So we are already told that segment AC is congruent to segment DC. They both have a right angle, as indicated by the angle symbol, and they share side-length BC.
According to the Hypotenuse-Leg Theorem, two right triangles that have a congruent hypotenuse and a corresponding, congruent leg are congruent triangles. AC and DC are hypotenuses and they are congruent. And BC, the shared side, is a corresponding congruent leg. And since they are both right triangles, we then know that the HL Theorem applies.
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)
