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Maru [420]
3 years ago
13

Suppose a possibly biased die is rolled 30 times and that the face containing

Engineering
1 answer:
WARRIOR [948]3 years ago
3 0

Answer:

p < α

0.01297 < 0.05

Since the p value is less than the α value therefore, we reject the null hypothesis so we have evidence to conclude that the die is biased.

Explanation:

H₀: The die is not biased

Ha: The die is biased

We can apply binomial distribution and determine whether the die is biased or not. (we can also perform z-test, it will provide similar results)

We know that a binomial distribution is given by  

P(x; n, p) = nCx pˣ (1 - p)ⁿ⁻ˣ  

Where p is the probability of success and 1 - p is the probability of failure, n is number of trials and x is the variable of interest.

For the given problem,

Total trials are n = 30

When you roll a die, there are total 6 possible outcomes,

The probability of getting the face containing two pips on each trial is

p = 1/6

p = 0.1667

The variable of interest is x = 10

P(10; 30, 0.1667) = ³⁰C₁₀*0.1667¹⁰*(1 - 0.1667)³⁰⁻¹⁰

P(10; 30, 0.1667) = (30045015)*(0.1667)¹⁰*(0.8333)²⁰

P(10; 30, 0.1667) = 0.01297

Assuming that the level of significance is α = 0.05 then

p < α

0.01297 < 0.05

Since the p value is less than the α value therefore, we reject the null hypothesis so we have evidence to conclude that the die is biased.

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Answer:

98,614.82 W/m²

Explanation:

Q = 2\pi hL(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})

Where;

Q = the amount of heat loss from the pipe

h =  the heat transfer coefficient of the pipe = 50 W/m².K

T₁ = the ambient temperature of the pipe = 30⁰C

T₂  = the outside temperature of the pipe = 100⁰C

L= the length of pipe

r₁ = inner radius of the pipe = 20mm

r₂ = outer radius of the pipe = 25mm

To determine the amount of heat loss from the pipe per unit length

From the equation above

\frac{Q}{L} = 2\pi h(\frac{T_2-T_1}{Ln\frac{r_2}{r_1}})

\frac{Q}{L} = 2\pi* 50(\frac{100-30}{Ln\frac{25}{20}})

\frac{Q}{L} = 314.159(\frac{70}{0.223})

\frac{Q}{L} = 314.159(313.901) = 98,614.82 W/m²

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Answer:

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A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027
Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = \frac{d}{2} = 3 inches

density,  \rho} = 493 lbm/ ft^{3}

S.G = 1.0027

g = 9.8 m/ m^{2} = 386.22 inch/ s^{2}

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where,

V = volume of sphere

C_{d} = drag coefficient

Now,

Surface area of sphere, A = 4\pi r^{2}

Volume of sphere, V = \frac{4}{3} \pi r^{3}

Using the above formulae in eqn (1):

v_{T} = \sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho  g}{4\pi r^{2} \rho C_{d}}}

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v_{T} = \frac{27.79}{\sqrt{C_d}} inch/sec

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