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garik1379 [7]
3 years ago
10

A drawer contains black socks and white socks. 40% of the socks are black socks. A number generator simulates randomly selecting

10 socks from the drawer. The number generator is used 10 times and the number of black socks in each trial is shown in the dot plot.
Which description is correct about the number generator being fair or not?




a)The number generator is not fair. It picked black socks half of the time.

b)The number generator is not fair. In three experiments, it picked black socks 50% of the time.

c)The number generator is fair. It picked the approximate percentage of black socks most of the time.

d)The number generator is fair. It picked black socks half of the time.

Mathematics
1 answer:
icang [17]3 years ago
8 0

Answer:

c)The number generator is fair. It picked the approximate percentage of black socks most of the time.


Step-by-step explanation:

given that 40% are black socks.

No of socks that can be black can take values as

X =0,1,2....10

Each sock is independent of the other.

Hence X is binomial with n = 10 , p = 0.4 and q = 0.6

P(x=r) = 10Cr (0.4)^r (0.6)^(10-r)

r = 0,1,2...10

P(X=0) = 0.0060

P(X=1) = 0.0403

P(X=2) = 0.1209


x                0       1             2             3             4              5             6          

Prob 0.0060 0.0403 0.1209 0.2150 0.2508 0.2007 0.1115


7          8           9               10

0.0425 0.0106 0.0016 0.0001

----

Comparing with the dot plot we find that the probability increases upto 4 attains peak and comes down and almost negligible for 0,1,9 and 10.

But for 6, 7, 8 prob is not 0 but approximately  dot plot is  consistent with the theoretical results.

c)The number generator is fair. It picked the approximate percentage of black socks most of the time.


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Initially 15 grams of salt are dissolved into 25 liters of water. Brine with concentration of salt 4 grams per liter is added at
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Answer:

a) dx/dt = 600 - 6x

b) x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

c) The mass of salt in the tank attains the value of 20 g at time, t = 0.227 min = 13.62 s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

Fᵢ = F₀ = F = 6 L/min

a) Then the component balance for the salt

Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

dx/dt = Fᵢxᵢ - Fx

Fᵢ = 6 L/min, C = 4 g/L, F = 6 L/min

dC/dt = 24 - 6C

dx/dt = 25 (dC/dt), (dC/dt) = (1/25) (dx/dt) and C = x/25

(1/25)(dx/dt) = 24 - (6/25)x

dx/dt = 600 - 6x

b) dC/dt = 24 - 6C

dC/(24 - 6C) = dt

∫ dC/(24 - 6C) = ∫ dt

(-1/6) In (24 - C) = t + k (k = constant of integration)

In (24 - 6C) = -6t - 6k

-6k = K

In (24 - 6C) = K - 6t

At t = 0, C = 15 g/25 L = 0.6 g/L

In (24 - 6(0.6)) = K

In 20.4 = K

K = 3.02

So, the equation describing concentration of salt at anytime in the tank is

In (24 - 6C) = K - 6t

In (24 - 6C) = 3.02 - 6t

24 - 6C = e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾

6C = 24 - (⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)

C = 4 - ((e⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)/6

C = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

But C = x/25

x/25 = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

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c) when x = 20 g

20 = 100 - 4.12(e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

80 = (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

- (6t - 3.02) = In 80

- (6t - 3.02) = 4.382

(6t - 3.02) = -4.382

6t = -4.382 + 3.02

t = 1.362/6 = 0.227 min = 13.62 s

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