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juin [17]
3 years ago
7

The table shows the number of feathers Patsy the Peacock sold at each of the 8 festivals this year.

Mathematics
2 answers:
Vlada [557]3 years ago
6 0

Answer:

75%

Step-by-step explanation:

It's right

ANTONII [103]3 years ago
4 0

We have been given a table that shows the number of feathers Patsy the Peacock sold at each of the 8 festivals this year.  We are asked to find the probability that Patsy sells fewer than 5 feathers next festival.

\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}}  

Number of favorable outcomes would be number of feathers less than 5.

We can see that there are 6 numbers less than 5 that are 3, 1, 4, 2, 3, 2.

The total number of years is 8. Upon substituting these values in probability formula, we will get:

\text{Probability of selling fewer than 5 feathers}=\frac{6}{8}

\text{Probability of selling fewer than 5 feathers}=\frac{3}{4}

\text{Probability of selling fewer than 5 feathers}=0.75

Let us convert 0.75 into percent.

0.75\times 100\%=75\%

Therefore, the probability that Patsy sells fewer than 5 feathers next year would be 75% and option D is the correct choice.

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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

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