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Vladimir79 [104]
3 years ago
9

A scientist measures a substance to be 0.8 grams. Calculate the percent of error in the measurement. Show all work for full cred

it.
Mathematics
1 answer:
Volgvan3 years ago
6 0
You'll need to give a bit more information for the question to be answered. You can only calculate the percentage of error if you know what the mass of the substance *should be* and what you've *measured* it to be.

In other words, if a substance has a mass of 0.55 grams and you measure it to be 0.80 grams, then the percent of error would be:

percent of error = { | measured value - actual value | / actual value } x 100%

So, in this case:

percent of error = { | 0.80 - 0.55 | / 0.55 } x 100%
percent of error = { | 0.25 | / 0.55 } x 100%
percent of error = 0.4545 x 100%
percent of error = 45.45%

So, in order to calculate the percent of error, you'll need to know what these two measurements are. Once you know these, plug them into the formula above and you should be all set!
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nika2105 [10]

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x^2-7x+12=0

To solve this equation, we can use the formula x^2-sx+p=0

It means that, if the leading terms is 1, then the x coefficient is the opposite of the sum of the roots, and the constant term is the product of the roots.

So, we're looking for two terms whose sum is 7, and whose product is 12. These numbers are easily found to be 3 and 4.

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2 years ago
What is 0.69696969 expressed as the quotient of two integers in simplest form?
timofeeve [1]

When written as the quotient of two integers,

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<u>If</u> you had said that the ' 69 ' keeps repeating forever and never ends,
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Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discha
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Answer:

Step-by-step explanation:

Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discharging hot water into the river is given by

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a. [0, 10]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

The average temperature

= (160 + 155)/2 = 157.5

b. [10, 40]

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (80 + 155)/2 = 117.5

c. [0, 40]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (160 + 80)/2 = 120

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Step-by-step explanation:

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