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2 years ago

If the probability of a win is 0.2 and the probability of a draw is 0.64, what is the probability of a loss?

Mathematics

2 answers:

maw [93]2 years ago

8 0

Answer:

0.16 or 16%

Step-by-step explanation:

an easier way to work this out is to convert these decimals into percentage. to do this, multiply the number by 100:

0.2 x 100 = 20%

0.64 x 100 = 64%

a percentage is out of 100 (probability of a win = 20/100 = 1/5) so to work this out, you just have to subtract the percentages that are known from 100:

100% - 20% = 80%

80% - 64% = 16%

∴ the probability of a loss = 16% or 0.16 as a decimal.

Another way to work this out is to add together the amounts that are given then work out what number needs to be added to = 1

0.2 + 0.64 = 0.84

0.84 + ? = 1

0.84 + 0.16 = 1

∴ the probability of a loss = 0.16

HOPE THIS WAS HELPFUL :-)

Veronika [31]2 years ago

5 0

To find the probability of a loss add .2 and .64 together. You will get .84. The remaining .16 is the probability of a loss. The probability of a loss is .16.

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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

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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

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

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

㏑0.01 = -t/100

t = -100㏑0.01

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Answer:

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Step-by-step explanation:

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