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3 years ago

The exchange rate is: £1 = $1.54 How many £ would I get for $77 Please help now

Mathematics

2 answers:

kirill [66]3 years ago

7 0

Answer:

Step-by-step explanation:

tiny-mole [99]3 years ago

4 0

Answer:

£50

Step-by-step explanation:

£1 = $1.54

£x = $77

Cross multiply.

1.54x = 77

x = 77/1.54

x = 50

You would get £50 for $77.

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spayn [35]

Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

What is the area of the figure ?

vodka [1.7K]

Answer:

48cm^2

Step-by-step explanation:

We can break the composite figure into two pieces.

We take the bottom rectangle and determine its area

The figure is 10 cm by 2 cm

A = 10*2 = 20cm^2

The height of the vertical rectangle is 9 cm -2 cm = 7 cm

The width is 10 cm - 6 cm = 4 cm

The figure is 7 cm cm by 4 cm

A = 7*4 = 28cm^2

The total area is the sum of the two areas

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5 0

3 years ago

The transformation of the triangle represents a translation.

seraphim [82]

The answer would be false.

This would not be a translation. Think of a translation as sliding a figure to a different position on a graph. This does not change how the shape is facing, but changes the location of it on the coordinate plane.

3 0

3 years ago

What are the prime factors of 700<br>A. 2×2×5×5×7<br>B. 350×2<br>C. 2×2×2×5×7<br>D. 2×2×25×7

Fudgin [204]

The correct answer is A
2×2×5×5×5×7
4 × 25 × 7
700

8 0

3 years ago

Compilers can have a profound impact on the performance of an application. Assume that for a program, compiler A results in a dy

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

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

8 0

3 years ago