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3 years ago

Poems that focus on one image usually have what purpose? PLEASE HELP MEH!!

Engineering

2 answers:

____ [38]3 years ago

5 0

Answer:

Poems that focus on one image usually have what purpose? PLEASE HELP MEH!!

<em>A. to make readers understand how one event leads to another</em>

B. to make readers look at something in a new and different way

C. to make a point about how two or three things are alike

Explanation:

the one is high lighted

Alinara [238K]3 years ago

5 0

Answer:

A. to make readers understand how one event leads to another

Explanation:

Hope this helped have an amazing day!

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On the first statistics exam, the coefficient of determination between the hours studied and the grade earned was 80%. The stand

True [87]

Answer:

$\left[\begin{array}{ccccc}&DF&SS&MS&F\\Regression&1&7200&7200&72\\Error&18&1800&100\\total&19&900\end{array}\right]$

Explanation:

Sample size, n=20

Degrees of freedom is 1

Number of degrees of freedom for error is n-2 hence 20-2=18

Total number of degrees of freedom is 18+1=19

Standard error estimate is $s_{y-x}=\sqrt {\frac {SSE}{n-2}}$

Here, $SSE=(n-2)s_{y-x}^{2}=(20-2)(10)^{2}=1800$

Coefficient of determination $r^{2}=\frac {SSE}{SS total}$

Here, $SSR=r^{2}(SSR+SSE)$

$SSR=\frac {r^{2}}{1-r^{2}} SSE=\frac {0.8}{1-0.8}(1800)=7200$

The total sum of squares is

SS total=SSR+SSE=7200+1800=9000

MSR=SSR=7200

$MSE=\frac {SSE}{n-2}=\frac {1800}{20-2}=100$

F value is given by

$F=\frac {MSR}{MSE}=\frac {7200}{100}=100$

The ANOVA table is then

$\left[\begin{array}{ccccc}&DF&SS&MS&F\\Regression&1&7200&7200&72\\Error&18&1800&100\\total&19&900\end{array}\right]$

4 0

4 years ago

provides steady-state operating data for a solar power plant that operates on a Rankine cycle with Refrigerant 134a as its worki

Vaselesa [24]

Answer:

hello some parts of your question is missing attached below is the missing part ( the required fig and table )

answer : The solar collector surface area = 7133 m^2

Explanation:

Given data :

Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2

percentage of solar power absorbed by refrigerant = 60%

Determine the solar collector surface area

The solar collector surface area = 7133 m^2

attached below is a detailed solution of the problem

8 0

3 years ago

What is the standard half-cell potential for the oxidation of methane under acidic conditions? The reaction for methane is as fo

Yuri [45]

Answer:

The element that is oxidized is carbon.

Its oxidation state increased. It increased from -4 to +4

Explanation:

Oxidation is a process that involves increase in oxidation number.

The oxidation number of carbon in CH4 is -4

C + (1×4) = 0

C + 4 = 0

C = 0 - 4 = -4

The oxidation number of carbon in CO2 is +4

C + (2×-2) = 0

C - 4 = 0

C = 0+4 = 4

Increase in the oxidation number of carbon from -4 to +4 means carbon is oxidized

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3 years ago

In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery

Dima020 [189]

Answer:

248.756 mV

49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

voltage: (100 V)(1000/(1000 +1000) = 50 V

impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

The open-circuit voltage is the difference between these terminal voltages:

(50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

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3 years ago

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3 years ago