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aksik [14]
3 years ago
12

There are many ways to produce crooked dice. To load a die so that 6 comes up too often and 1 , which is opposite 6 , comes up t

oo seldom, add a bit of lead to the filling of the spot on the 1 face. If a die is loaded so that 6 comes up with probability 0.2 and the probabilities of the 2,3,4 , and 5 faces are not affected, which of the choices is the correct assignment of probabilities to the six faces

Mathematics
1 answer:
Andreas93 [3]3 years ago
4 0

Answer:

1            2                  3             4            5       6

2 ÷ 15   1 ÷ 6           1 ÷ 6        1 ÷ 6     1 ÷ 6    1 ÷ 5

Step-by-step explanation:

As it is mentioned in the question that the probability of faces like 2,3,4,5 are not affected so in this case

P(2) = P(3) = P(4) = P(5) = 1 ÷ 6

And,  P(6) = 0.2

As we know that

P(1) + P(2) +P(3) +P(4) +P(5) +P(6) = 1

This results P(1) = 2 ÷ 15

So,

1            2                  3             4            5       6

2 ÷ 15   1 ÷ 6           1 ÷ 6        1 ÷ 6     1 ÷ 6    1 ÷ 5

Hence, the correct option is 2nd

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Answer:

A familiar situation is: cost of books you pay for versus the quantity of books bought.

Cost of books ($) and quantity of books are directly proportionally related in the situation.

The graph will look like the graph in the attachment below.

A quantity (dependent variable) will change constantly in relation to another quantity (independent variable) if the relation is a proportional relationship.

A familiar situation for example can be the cost you pay for books will be directly proportional or dependent on the number of books you bought.

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A change in the number of books will cause a change in the cost you will pay for buying books.

This shows a direct proportional relationship between the two quantities.

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A familiar situation is: cost of books you pay for versus the quantity of books bought.

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Step-by-step explanation:

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3 years ago
A triangle has one acute angle as the complement of the other. find the third angle​
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3 years ago
A tank contains 100 L of water. A solution with a salt con- centration of 0.4 kg/L is added at a rate of 5 L/min. The solution i
Fantom [35]

Answer:

a) (dy/dt) = 2 - [3y/(100 + 2t)]

b) The solved differential equation gives

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

c) Concentration of salt in the tank after 20 minutes = 0.2275 kg/L

Step-by-step explanation:

First of, we take the overall balance for the system,

Let V = volume of solution in the tank at any time

The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)

The rate of change of the volume of solution = dV/dt

Rate of flow into the tank = Fᵢ = 5 L/min

Rate of flow out of the tank = F = 3 L/min

(dV/dt) = Fᵢ - F

(dV/dt) = (Fᵢ - F)

dV = (Fᵢ - F) dt

∫ dV = ∫ (Fᵢ - F) dt

Integrating the left hand side from 100 litres (initial volume) to V and the right hand side from 0 to t

V - 100 = (Fᵢ - F)t

V = 100 + (5 - 3)t

V = 100 + (2) t

V = (100 + 2t) L

Component balance for the amount of salt in the tank.

Let the initial amount of salt in the tank be y₀ = 0 kg

Let the rate of flow of the amount of salt coming into the tank = yᵢ = 0.4 kg/L × 5 L/min = 2 kg/min

Amount of salt in the tank, at any time = y kg

Concentration of salt in the tank at any time = (y/V) kg/L

Recall that V is the volume of water in the tank. V = 100 + 2t

Rate at which that amount of salt is leaving the tank = 3 L/min × (y/V) kg/L = (3y/V) kg/min

Rate of Change in the amount of salt in the tank = (Rate of flow of salt into the tank) - (Rate of flow of salt out of the tank)

(dy/dt) = 2 - (3y/V)

(dy/dt) = 2 - [3y/(100 + 2t)]

To solve this differential equation, it is done in the attached image to this question.

The solution of the differential equation is

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

c) Concentration after 20 minutes.

After 20 minutes, volume of water in tank will be

V(t) = 100 + 2t

V(20) = 100 + 2(20) = 140 L

Amount of salt in the tank after 20 minutes gives

y(t) = 0.4 (100 + 2t) - 40000 (100 + 2t)⁻¹•⁵

y(20) = 0.4 [100 + 2(20)] - 40000 [100 + 2(20)]⁻¹•⁵

y(20) = 0.4 [100 + 40] - 40000 [100 + 40]⁻¹•⁵

y(20) = 0.4 [140] - 40000 [140]⁻¹•⁵

y(20) = 56 - 24.15 = 31.85 kg

Amount of salt in the tank after 20 minutes = 31.85 kg

Volume of water in the tank after 20 minutes = 140 L

Concentration of salt in the tank after 20 minutes = (31.85/140) = 0.2275 kg/L

Hope this Helps!!!

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A pair of shoes is on sale for 15% off. With this discount, customers will save
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$9 is the part, 15% is the percent, and $60 is the whole

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