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andrey2020 [161]
2 years ago
11

Charlie has the utility function u(xa, xb) =xaxb.his indifference curve passing through 32 applesand 8 bananas will also pass th

rough the point where he consumes 4 apples and
Mathematics
1 answer:
natka813 [3]2 years ago
4 0
<span>On an indifference curve, all bundles give the same amount of utility. (32,8) gives a utility of U(32,8)=32x8=256 If (4,y) is on the same indifference curve, then it must give the same utility. Hence, 256 = 4y y=64 64 bananas</span>
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School meeting was attended only by sophomores, juniors, and seniors. 5/12 of those who attended were juniors, and 1/3 were seni
Veronika [31]
If \frac{9}{12} of students were juniors and seniors that leaves the 36 sophomores to be \frac{3}{12} or \frac{1}{4} of attendees. I got \frac{9}{12} by converting \frac{1}{3} to \frac{4}{12} by multiplying it by 4 and adding it to \frac{5}{12}.

Now if 36 is \frac{3}{12} of the equation that means theirs at least 36 juniors and seniors.

Lets start with seniors. there should be just over 36 of them there. Lets start by multiplying 36 by \frac{3}{12} to figure out how many \frac{1}{12} is.
The equation would look like this 36*(3/12)
start with 3/12. this equals .25 or \frac{1}{4}.
Now multiply 36 by .25. this equals 9.

We have 9 students per \frac{1}{12} of the attendees.

To get the amount of seniors add 9 to 36 \frac{4}{12} because we already stated \frac{3}{12} is 36.
36+9=45

Now to find the juniors add another 9 to create \frac{5}{12}.
45+9=54

In total you have:
36 sophomores
54 juniors
45 seniors

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3 0
2 years ago
The Greens want to put an addition on their house 18 months from now. They will need to save $10,620 in order to achieve this go
ValentinkaMS [17]
Option A. is the correct answer :)
4 0
2 years ago
Find the exponential generating function for the number of alphanumeric strings of length n n formed from the 26 26 uppercase le
lana66690 [7]

Answer:

the answer is contained in the attachment

Step-by-step explanation:

for complete explanation kindly check the attachment.

5 0
2 years ago
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
2 years ago
An account is opened with $7,595.96 with a rate of increase of 2% per year. After 1 year, the bank account contains $7,746.90. A
wolverine [178]

Answer:

y = 7,595.96(1.02)x

Step-by-step explanation:

y = 7,595.96(1.02)x

a normal linear equation is basically y = x + b

in this case,

  • y = amount of money in the bank account
  • x = number of years
  • b is not required because there are no new deposits or withdrawals from the account

Since the accounts earns 2% interest per year, then you need to multiply the original amount by 1.02. As more years pass, the account will increase by 1.02, so the account increases by 1.02x. Since the original amount is $7,595.96, that will be our starting point.

3 0
2 years ago
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