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mylen [45]
2 years ago
6

There are 2.54 centimeters in 1 inch. There are 100 centimeters in 1 meter.

Mathematics
2 answers:
DochEvi [55]2 years ago
3 0

Answer:

About 4.06, so your answer is 4 meters.

soldi70 [24.7K]2 years ago
3 0

Answer:

406.4

Step-by-step explanation:

.  160 x 2.54 = 406.4 meters in 160 inches.

.  To check you can do division.

.  406.4 divided by 2.54 = 160 inches.

Hope this help!

Have a great day!

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S/4 -7 =9<br> any help solving this question?
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Answer:

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Step-by-step explanation:

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Determine the standard variation of the data below. (1, 2, 3, 4, 5)
Ne4ueva [31]
Calculate for the mean/ average of the given numbers:
 
                             μ = (1 + 2 + 3 + 4 + 5) / 5 = 3

Then, we calculate for the summation of the squares of differences of these numbers from the mean, S
 
                             S  = (1 - 3)² + (2 - 3)² + (3 - 3)² + (4 - 3)² + (5 - 3)²
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Divide this summation by the number of items and take the square root of the result to get the standard deviation.

                              SD = sqrt (10 / 5) = sqrt 2  
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Thus, the standard deviation of the given is equal to 1.41. 
              
                            
8 0
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How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy
lisov135 [29]

Answer:

300\; \rm lb.

Step-by-step explanation:

Let x represent the mass (in pounds) of that 15\% copper alloy required, such that the final mixture would contain 25.5\% copper by mass.

Consider: if x pounds of that 15\% copper alloy is mixed with 700 pounds that 30\% copper alloy, what would be the mass of copper in the mixture?

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Therefore, the mixture would contain (210 + 0.15\, x) \; \rm lb of copper.

The mass of that mixture would be (700 + x)\; \rm lb. The mass fraction of copper in that mixture would be:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%.

This ratio is supposed to be equal to 25.5\%. These two pieces of equations combine to give an equation about x:

\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%.

\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255.

Simplify and solve for x:

210 + 0.15\, x= 0.255\, (700 + x).

(0.255 - 0.15)\, x= 210 - 0.255 \times 700.

\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300.

Therefore, 300\; \rm lb of that 15\% alloy would be required.

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