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3 years ago

9.6 Code practice Edhesive

Computers and Technology

1 answer:

Oxana [17]3 years ago

7 0

Answer:

N = [1,1,1,1,1],

[2,2,2,2,2],

[3,3,3,3,3],

[4,4,4,4,4]

def printIt(ar):

for row in range(len(ar)):

for col in range(len(ar[0])):

print(ar[row][col], end=" ")

print("")

N=[]

for r in range(4):

N.append([])

for r in range(len(N)):

value=1

for c in range(5):

N[r].append(value)

value=value + 1

printIt(N)

print("")

newValue=1

for r in range (len(N)):

for c in range(len(N[0])):

N[r][c] = newValue

newValue = newValue + 1

printIt(N)

Explanation:

I got 100%.

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Answer:

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Explanation:

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$E = CV^2f$

Where, $E$ is the energy

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$V$ is the voltage and

$f$ is the clock rate.

Let $E_1$ be the energy of older processor.

$C_1$ be the capacitance of older processor.

$V_1$ be the voltage of older processor

$f_1$ be the capacitance of older processor

So, $E_1 = C_1V_1^2f_1$ ....... (1)

and

Let $E_2$ be the energy of newer processor.

$C_2$ be the capacitance of newer processor.

$V_2$ be the voltage of newer processor

$f_2$ be the capacitance of newer processor

$E_2 = C_2V_2^2f_2$ ....... (2)

Dividing equation (2) with equation (1):

$\dfrac{E_2}{E_1} = \dfrac{C_2V_2^2f_2}{C_1V_1^2f_1}$

As per given statement:

$C_1=C_2$

$V_2=1.2\times V_1$

$f_2=1.2\times f_1$

Putting the values above:

$\dfrac{E_2}{E_1} = \dfrac{C_1\times (1.2V_1)^2\times 1.2f_1}{C_1V_1^2f_1}\\\Rightarrow \dfrac{E_2}{E_1} = \dfrac{(1.2)^2\times 1.2}{1}\\\Rightarrow \dfrac{E_2}{E_1} = 1.728\\\Rightarrow \bold{E_2 = 1.728 \times E_1}$

Energy consumed by newer processor is <em>1.728 </em>times the energy consumed by older processor.

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