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Marina CMI [18]

2 years ago

6

Raina wants to paint the ceiling of her restaurant. The ceiling is in the shape of a square. Its side lengths are 55 feet. Suppo

se each can of paint will cover 275 square feet. How many cans will she need to paint the ceiling?

1 answer:

Ksju [112]2 years ago

8 0

area of the ceiling = 55 x 55 = 3025 square feet

3025/275 = 11 cans

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Determine a series of transformations that would map polygon ABCD onto polygon A’B’C’D’?

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Answer:

A Translation 4 units leftwards followed by a reflection across the x-axis.

Step-by-step explanation:

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Translation 4 units leftwards.

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Answer:

a) $y=9x$

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Where k is the constant of proportionality.

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b) Solve for k:

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mixas84 [53]

Incomplete question as we have not told what to find.So I have assumed to find average CPI for each program.So complete question is here

Compilers can have a profound impact on the performance of an application. Assume that for a program, compiler A results in a dynamic instruction count of 1.0E9 and has an execution time of 1.1 s, while compiler B results in a dynamic instruction count of 1.2E9 and an execution time of 1.5 s.

Find the average CPI for each program given that the processor has a clock cycle time of 1 ns.

Answer:

$CPI_{A}=1$

$CPI_{B}=1.25$

Step-by-step explanation:

As we know CPU time given as

$CPU_{time}=instructions*CPI*Cycle_{time} \\So\\CPI=\frac{CPU_{time}}{instructions*Cycle_{time}}\\ Where\\Cycle_{time}=1*10^{-9}s\\ And\\executionTime=CPUtime$

For Compiler A

$CPI_{A}=\frac{CPU_{timeA} }{instruction_{A}*Cycle_{time} }\\CPI_{A}=\frac{1s}{10^{9}*1.0*10^{-9}s }\\CPI_{A}=1$

For Compiler B

$CPI_{B}=\frac{CPU_{timeB} }{instruction_{B}*Cycle_{timeB} }\\CPI_{B}=\frac{1.5s}{(1.2*10^{9})*(1.0*10^{-9}s) }\\CPI_{B}=1.25$

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