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Aleks04 [339]
4 years ago
14

A _________ is interesting only if the statistics computed from transactions covered by the rule are different than those comput

ed from transactions not covered by the rule.
Engineering
1 answer:
qwelly [4]4 years ago
7 0

Answer: Quantitative association rule.

Explanation:

The quantitative rules of association apply to the basic type of rules of association which exists as X and Y, with X and Y consisting of a collection of numerical and/or categorical attributes. Unlike general association laws, where both the left and right sides of the law should have categorical (nominal or discrete) attributes, a numerical attribute must be included in at least one attribute of the quantitative association rule (left or right).

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A series circuit has 4 identical lamps. The potential difference of the energy source is 60V. The total resistance of the lamps
Alexxx [7]

Answer:

I=3A

Explanation:

From the question we are told that:

Number of lamps N=4

Potential difference V=60v

Total Resistance of the lamp is R= 20ohms

Generally the equation for Current I is mathematically given by

 I=\frac{V}{R}

 I=\frac{60}{20}

 I=3A

8 0
3 years ago
An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a
DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π ×  0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶  × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

4 0
3 years ago
PLEASE ANSWER ASAP 20 points and brainlyyy
k0ka [10]

Answer:

top right

Explanation:

4 0
3 years ago
Read 2 more answers
We have a credit charge that is trying to process but we do not remember signing up and email login is not working? Is there a w
Dmitry_Shevchenko [17]

Answer:

<u>Yes</u>

Explanation:

In such a case, one way to check the <em>credit charge</em> is to <u>contact your bank, </u>doing so would allow the bank to check your account properly to determine where the transaction was originated from.

Another way you could check is to contact the online merchant where such a transaction was initiated.

6 0
3 years ago
The volume of 1.5 kg of helium in a frictionless piston-cylinder device is initially 6 m3. Now, helium is compressed to 2 m3 whi
coldgirl [10]

Answer:

The initial temperature will be "385.1°K" as well as final will be "128.3°K".

Explanation:

The given values are:

Helium's initial volume, v₁ = 6 m³

Mass, m = 1.5 kg

Final volume, v₂ = 2 m³

Pressure, P = 200 kPa

As we know,

Work, W=p(v_{2}-v_{1})

On putting the estimated values, we get

⇒            =200000(2-6)

⇒            =200000\times (-4)

⇒            =800,000 \ N.m

Now,

Gas ideal equation will be:

⇒  pv_{1}=mRT_{1}

On putting the values. we get

⇒  200000\times 6=1.5\times 2077\times T_{1}

⇒  T_{1}=\frac{1200000}{3115.5}

⇒       =385.1^{\circ}K (Initial temperature of helium)

and,

⇒  pv_{2}=mRT_{2}

On putting the values, we get

⇒  200000\times 2=1.5\times 2077\times T_{2}

⇒  T_{2}=\frac{400000}{3115.5}

⇒       =128.3^{\circ}K (Final temperature of helium)

3 0
3 years ago
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