8/15 or 11/15 would b the probabiltity of those
2t+2d+2c=30
2(8.50)+2(3.50)+2(3.00)=$30
t representing tickets d representing drinks c representing candy
the answer really depends on how much of each item they buy
if they bought more than 2 of each just change the 2 to the number of items bought
We will use substitution to solve this system of linear equations, as the first equation has x and y with no coefficients, which makes it easier to find one in terms of the other. We can then substitute that value in the other equation and find the values of x and y.
x = y + 5 ---> equation 1
3x + 2y = 5 ---> equation 2
From equation 1, we get the value of x as y + 5. Using the substitution method, we can find the value of y by substituting (y+5) for x in the 2nd equation.
3(y+5) + 2y = 5
3y + 15 + 2y = 5
5y = 5 - 15
5y = -10
y = -2
Subsituting this value of y in (y+5), we can find x.
x = y + 5
x = -2 + 5
x = 3
Therefore, x = 3 and y = -2.
I will also solve this using elimination method.
Let us multiply equation 1 by 2, so that we get 2y in both equations.
2x = 2y + 10
3x + 2y = 5
Let us add both the equations.
2x + 3x + 2y = 5 + 2y + 10
5x = 15 + 2y - 2y
5x = 15
x = 3
Substituting this value of x in equation 1, we get
x = y + 5
3 = y + 5
y = 3 - 5
y = -2
Therefore, x = 3 and y = -2.
9514 1404 393
Answer:
(dN)/(dt) = (0.4)/(1200)N(1200-N) -50
142 fish
Step-by-step explanation:
A) The differential equation is modified by adding a -50 fish per year constant term:
(dN)/(dt) = (0.4)/(1200)N(1200-N) -50
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B) The steady-state value of the fish population will be when N reaches the value that makes dN/dt = 0.
(0.4/1200)(N)(1200-N) -50 = 0
N(N-1200) = -(50)(1200)/0.4) . . . . rewrite so N^2 has a positive coefficient
N^2 -1200N + 600^2 = -150,000 +600^2 . . . . complete the square
(N -600)^2 = 210,000 . . . . . simplify
N = 600 + √210,000 ≈ 1058
This steady-state number of fish is ...
1200 - 1058 = 142 . . . . below the original carrying capacity
Answer:
$1064.20
Step-by-step explanation:
36 payments of $168.45 have a total of $6064.20. The excess over the loan amount of $5000 is the interest paid:
$6064.20 -5000 = $1064.20 . . . . interest paid
