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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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The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
PLS HELP ME
Oksana_A [137]

Answer:

The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.

Explanation:

160 - 120 = 40

120 = 100

40 = X

40 x 100 / 120 = X

4000 / 120 = X

33.333 = X

120 = 100

160 = X

160 x 100 /120 = X

16000 / 120 = X

133.333 = X

4 0
3 years ago
2.) A fluid moves in a steady manner between two sections in a flow
Talja [164]

Answer:

250\ \text{lbm/min}

625\ \text{ft/min}

Explanation:

A_1 = Area of section 1 = 10\ \text{ft}^2

V_1 = Velocity of water at section 1 = 100 ft/min

v_1 = Specific volume at section 1 = 4\ \text{ft}^3/\text{lbm}

\rho = Density of fluid = 0.2\ \text{lb/ft}^3

A_2 = Area of section 2 = 2\ \text{ft}^2

Mass flow rate is given by

m=\rho A_1V_1=\dfrac{A_1V_1}{v_1}\\\Rightarrow m=\dfrac{10\times 100}{4}\\\Rightarrow m=250\ \text{lbm/min}

The mass flow rate through the pipe is 250\ \text{lbm/min}

As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

m=\rho A_2V_2\\\Rightarrow V_2=\dfrac{m}{\rho A_2}\\\Rightarrow V_2=\dfrac{250}{0.2\times 2}\\\Rightarrow V_2=625\ \text{ft/min}

The speed at section 2 is 625\ \text{ft/min}.

3 0
3 years ago
Match the military operation to the category of satellite that would perform it.
SOVA2 [1]

Answer:

1. Location of enemy ground troops  - EARTH OBSERVING.

Using earth observing satellite imagery, the military can observe vast expanses of land and in so doing, find the location of enemy ground troops.

2. Routine reconnaissance of an unfamiliar climate  - WEATHER

In other to find out more about the climate of an area, a weather satellite can be used to observe the areas and its changing weather patterns.

3. Analyze waterways in an unfamiliar location  - NAVIGATION

Using navigation satellites, navigation conduits such as roads and waterways can be observed.

4. Provide warning of an attack - COMMUNICATION.

Communications satellites enable people to communicate over great distances and so can be used by the military to warn of an impending attack.

5 0
3 years ago
Ben İngiliz oldum düzelte bilirmiyim
Lelu [443]

Answer:

What laguange is that?

Explanation:

7 0
3 years ago
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