<span>opposite sides and _____ </span>and angles
Answer:
2.45c + 1.65c = 4.12 + 0.75
Step-by-step explanation:
To write an equation to find the value for c, we need to declare what c is first.
c = price of fruit
2.45c + 1.65c = 4.12 + 0.75
Now we multiplied c to 2.45 and 1.65 and added them together, because whatever the value of c is will give us the equivalence of the sum of 4.12 + 0.75.
Now to check if the equation is right, let's solve for c.
2.45c + 1.65c = 4.12 + 0.75
4.1c = 4.87
Now to get the value of c, we divide both sides of the equation by 4.1.
c = 1.19
Now let's substitute the value of c in the equation to see if we got it right.
2.45(1.19) + 1.65(1.19) = 4.12 + 0.75
2.92 + 1.96 = 4.87
4.87 = 4.87
Therefore concluding that the value of c is 1.19.
Since the sum of all probabilities of all all elementary events will always be equal to 1. Furthermore, the probabilities of all mutually exclusive set of events that is part of the entire sample space will always be total of 1.
So in the problem, the answer is 1/8.
1/8 for red + 3/8 for green + 3/8 for yellow + 1/8 for blue = 8/8 or 1.
<span>Let L be the number of yards on a roll of lace ribbon.
Let S be the number of yards on a roll of satin ribbon.
We can set up two equations.
equation 1: 3L + 2S = 120 yards
equation 2: 2L + 4S = 160 yards
We can multiply (equation 1) by 2 and subtract (equation 2).
equation 1: 6L + 4S = 240 yards
equation 2: 2L + 4S = 160 yards
4L = 80 yards
L = 20 yards
equation 1: 3L + 2S = 120 yards
3(20 yards) + 2S = 120 yards
2S = 60 yards
S = 30 yards
There are 20 yards on a roll of lace ribbon.
There are 30 yards on a roll of satin ribbon.</span>
Answer:
a.) C(q) = -(1/4)*q^3 + 3q^2 - 12q + OH b.) $170
Step-by-step explanation:
(a) Marginal cost is defined as the decrease or increase in total production cost if output is increased by one more unit. Mathematically:
Marginal cost (MC) = change in total cost/change in quantity
Therefore, to derive the equation for total production cost, we need to integrate the equation of marginal cost with respect to quantity. Thus:
Total cost (C) = Integral [3(q-4)^2] dq = -(1/4)*(q-4)^3 + k
where k is a constant.
The overhead (OH) = C(0) = -(1/4)*(0-4)^3 + k = -16 + k
C(q) = -(1/4)*(q^3 - 12q^2 + 48q - 64) + k = -(1/4)*q^3 + 3q^2 - 12q -16 + k
Thus:
C(q) = -(1/4)*q^3 + 3q^2 - 12q + OH
(b) C(14) = -(1/4)*14^3 + 3*14^2 - 12*14 + 436 = -686 + 588 - 168 + 436 = $170
