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sweet [91]
3 years ago
7

A claw arcade machine contains stuffed animals and mystery boxes. There are 51 items in the machine. Winning a stuffed animal ha

s twice as many favorable outcomes as winning a mystery box. How many of each item is in the machine?
Mathematics
1 answer:
Oliga [24]3 years ago
4 0
Answer: 34 stuffed animals & 17 mystery boxes.

First off, split 51 into three parts, trying to find how much 1/3 is of 51. It equals 17.

The stuffed animals in the machine are “twice as much” of the mystery boxes, so you need to multiply 17 x 2 to find the amount of stuffed animals.

17 x 2 = 34

As for the mystery boxes, you need to multiply it by one.

17 x 1 = 17.

In summary, the answer is 34 stuffed animals and 17 mystery boxes.
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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
3 years ago
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