Answer:
I got to say Post Malone.
Step-by-step explanation:
1. He isn’t afraid to be relatable:
2. He brings his quirks to his Interviews:
3. His music breaks multiple genres:
4. The success of his first album:
7. He’s only 22:
Answer:
The correct answer would be , a = 168 tickets were sold
Step-by-step explanation:
Cost of adults ticket = $10.5
Cost of Students Ticket = %3.75
Sales Totals for opening night = $2071.5
Equation would be
10.5a + 3.75b = 2071.5
If 82 students attended the the play at that night, it means the value of b would be 82. Thus substituting the value in the above equation, we can find the value of a, which means how many adult tickets were sold.
10.5a + 3.75(82) = 2071.5
10.5a + 307.5 = 2071.5
10.5a = 2071.5-307.5
10.5a = 1764
a= 168 tickets were sold.
Answer:
its five
if you want a explanation, lmk
Answer:
if. x=2
y=7-x
y=7-2
y=2
Answer y=2
Answer:
a) The probability is 0.04
b) The probability is 0.36
c) The pprobability is 0,25
d) The probability is 0.09
Step-by-step explanation:
Lets calculate areas:
the target has a radius of 10 inces, hence the target area has a area on 10²*π = 100π square inches.
a) A circle of 2 inches of radius has an area of 2²π = 4π square inches, hence the probability of hitting that area is 4π/100π = 1/25 = 0.04
b) If the dart s within 2 inches of the rim, then it is not at distance 8 inches from the center (that is the complementary event). The probability for the dart to be at 8 inches of the center is 8²π/100π = 64/100 = 16/25 = 0.64, thus, the probability that the dart is at distance 2 or less from the rim is 1-0.64 = 0.36.
c) The first quadrant has an area exactly 4 times smaller than the area of the target (each quadrant has equal area), thus the probability for the dart to fall there is 1/4 = 0.25
d) If the dart is within 2 inches from the rim (which has probability 0.36 as we previously computed), then it will be equally likely for the dart to be in either of the 4 quadrants (the area that is within 2 inches from the rim forms a ring and it has equal area restricted on each quadrant). Therefore, the probability for the dice to be in the first qudrant and within 2 inches from the rim is 0.36*1/4 = 0.09.