1. Y = 10x
2. Y = 100 + 30x
3. Y = 50 + 1x
5. Y = 15 + .12x
If triangle IUP has angles I=50, U=60, P=70. The longest side of the triangle would be IU because if you draw the triangle and put the amounts of the angles in the correct place you draw a line across from the biggest angle to the side across from it and that gives you the longest side
(a) The probability that there is no open route from A to B is (0.2)^3 = 0.008.
Therefore the probability that at least one route is open from A to B is given by: 1 - 0.008 = 0.992.
The probability that there is no open route from B to C is (0.2)^2 = 0.04.
Therefore the probability that at least one route is open from B to C is given by:
1 - 0.04 = 0.96.
The probability that at least one route is open from A to C is:

(b)
α The probability that at least one route is open from A to B would become 0.9984. The probability in (a) will become:

β The probability that at least one route is open from B to C would become 0.992. The probability in (a) will become:

Gamma: The probability that a highway between A and C will not be blocked in rush hour is 0.8. We need to find the probability that there is at least one route open from A to C using either a route A to B to C, or the route A to C direct. This is found by using the formula:


Therefore building a highway direct from A to C gives the highest probability that there is at least one route open from A to C.
( a ) The system of equations:
2 r + b = 8.403 r + b = 9.35( b ) Graph is in the attachment.
( c ) Each item costs:
b = $6.50, r = $0.95We can prove it: 2 * 0.95 + 6.50 = 8.40
3* 0.95 + 6.50 = 9.35
Answer:
That would be sina.
Step-by-step explanation:
sin(a+b) = sinacosb + cosasinb
sin(a-b) = sinacosb - cosasinb
Adding we get sin(a+b) + sin(a-b) = 2sinaccosb
so sinacosb = 1/2sin(a+b) + sin(a-b)