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klasskru [66]
3 years ago
13

In Example 2-1, part c, the data were represented by the normal distribution function f(x)=0.178 exp(-0.100(x-451)2 Use this dis

tribution function to determine the fraction of individuals demonstrating a response in the range of 2.5 to 7.5.
Engineering
1 answer:
valkas [14]3 years ago
6 0

Answer:

P ( 2.5 < X < 7.5 ) = 0.7251

Explanation:

Given:

- The pmf for normal distribution for random variable x is given:

                                      f(x)=0.178 exp(-0.100(x-4.51)^2)

Find:

the fraction of individuals demonstrating a response in the range of 2.5 to 7.5.

Solution:

- The random variable X follows a normal distribution with mean u = 4.51, and standard deviation s.d as follows:

                               s.d = sqrt ( 1 / 0.1*2)

                               s.d = sqrt(5) =2.236067

- Hence, the normal distribution is as follows:

                               X ~ N(4.51 , 2.236)

- Compute the Z-score values of the end points 2.5 and 7.5:

              P ( (2.5 - 4.51) / 2.236 < Z < (7.5 - 4.51 ) / 2.236 )

              P ( -0.898899327 < Z < 1.337168651 )  

- Use the Z-Table for the probability required:

              P ( 2.5 < X < 7.5 ) = P ( -0.898899327 < Z < 1.337168651 ) = 0.7251            

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A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat
Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ

b) The coefficient of performance is:

COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6

c) The coefficient of performance of a reversible heat pump is:

COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l}  }

Th=20+273=293 K

Tl=10+273=283K

Replacing:

COP_{reversible}=\frac{293}{293-283}=29.3

4 0
3 years ago
According to the amortization table, Demarco and Tanya will pay a total of in interest over the life of their loan.
Ymorist [56]

Answer:

(Interest rate/number of payments)*$170000= interest for the first month.

Interest amounts for all the months of repayment plus $170000=Total loan cost

Explanation:

Interest is the amount you pay for taking a loan from a bank on top of the original amount borrowed.

Factors affecting how much interest is paid are; the principal amount, the loan terms, repayment schedule, the repayment amount and the rate of interest.

The interest paid=(rate of interest/number of payments to make)*principal amount borrowed.

You divide the interest with number of payments done in a year where monthly are divided by 12.Multiplying it by loan balance in the first month which is your principal amount gives the interest rate to pay for that month.

You new loan balance will be= Principal -(repayment-interest)

Do this for the period the loan should take.

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8 0
2 years ago
Read 2 more answers
Engineering Questions
valkas [14]
5 is the correct one to choose for this
6 0
2 years ago
Read 2 more answers
A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g
SSSSS [86.1K]

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

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