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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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3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the
gayaneshka [121]

Answer:

x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

                          V_fat = x_fat*V

- The volume of the muscle is given by:

                          V_muscle = V - V_fat

                                            = V - x_fat*V

                                            =V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

                         mass_fat + mass_muscle = Total average mass

                         p_fat*V_fat + p_muscle*V_muscle = p_avg*V

                         p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

                         Weight reading (Wsa) = m = p_1*V

                         p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

                         Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

                         Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

                         Where, p_2 = p_1 - p_water

                         p_2 = Wsw / V

- The average density p_avg:

                         p_avg = 0.5*(p_1 + p_2)  

                         p_avg = 0.5*(Wsa / V + Wsw / V)  

                         p_avg = 0.5*(Wsa + Wsw) / V                      

- Plug in the mass equation:

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

                         x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

                   x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

                         

6 0
3 years ago
A. The ragion was colonized by European powers
alex41 [277]

Answer:

?

Explanation:

3 0
2 years ago
In normal operation, a paper mill generates excess steam at 20 bar and 400◦C. It is planned to use this steam as the feed to a t
Keith_Richards [23]

Answer:

The maximum power that can be generated is 127.788 kW

Explanation:

Using the steam table

Enthalpy at 20 bar = 2799 kJ/kg

Enthalpy at 2 bar = 2707 kJ/kg

Change in enthalpy = 2799 - 2707 = 92 kJ/kg

Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s

Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW

6 0
3 years ago
Technician A says test lights are great for performing simple tests. Technician B says you can use a test light to check SRS cir
adoni [48]

The technician that is correct about either testing lights for simple tests or to check SRS Circuits is; Technician A.

<h3>Which Technician is Correct?</h3>

First of all it is pertinent to note that test lights are generally small bulbs that are turned on by the voltage and current flowing through the circuit in analog circuits.

Now, the  two values ​​of voltage and current are high and sufficient to light up the bulb. However, in digital circuits, the current is very small in the order of milliamps, and as a result there is not enough power to turn on the lights.

Thus, we can conclude that Technician A is correct.

Read more about Correct Technician at; brainly.com/question/14449935

5 0
2 years ago
To increase the thermal efficiency of a reversible power cycle operating between thermal reservoirs at TH and Tc, would you incr
alukav5142 [94]

<u></u>\ T_{c} has greater effect.

<u>Explanation</u>:

\eta_{\max }=1-\frac{T_{c}}{T_{A}}

T_{c}\\ = Temperature of cold reservoir

T_{H} = Temperature of hot reservoir

when T_{c} is decreased by 't',

$\eta_{\text {incre }}$ = 1-\frac{\left(\tau_{c}-t\right)}{T_{H}}

=n \ + \frac{t}{T_{n}}      -(i)

when {T_{H}} is increased by 'T'

\eta_{i n c}=\frac{n+\frac{t}{T_{H}}}{\left(1+\frac{k}{T_{H}}\right)}-(ii)

\eta_{\text {incre }} \ T_{c}>\eta_{\text {incre }} T_{\text {H }}

7 0
3 years ago
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