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qwelly [4]
3 years ago
8

4.890 in expanded form?

Mathematics
1 answer:
dangina [55]3 years ago
6 0

Answer: The student’s values are accurate as well as precise.

Explanation:

Precision refers to the closeness of two or more measurements to each other.

For Example: If you weigh a given substance three times and you get same value each time. Then the measurement is very precise.

Accuracy refers to the closeness of a measured value to a standard or known value.

For Example: If the mass of a substance is 50 kg and one person weighed 49 kg and another person weighed 48 kg. Then, the weight measured by first person is more accurate.

Given: Mass = 5.000 g

Mass weighed by A has values 4.891 g , 4.901 g and 4.890. Thus the average value is

Thus as the measured value is close to the true value, the student’s values are accurate and as the values are close to each other, the measurement is precise.

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A battery was charged. When the changing began, it was 23 percent full. After 30 minutes of charging, the battery was 89 percent
Aleks04 [339]
Before it charged, the battery was at 23%. In half and hour, it went to 89%. So, it rose 89-23, or 66% in half an hour. Now that we know how much it rose and that there are 30 mins in half an hour, we have to divide 66% by 30 to figure out the answer.

66/30 is 2.2. 

So, the battery rose at 2.2% per minute. 

Hope this helps!
4 0
4 years ago
If k and b are constant such that lim x approach infinity (kx+b-(x^3+1)/(x^2+1)=0. Find the values of k and b
Anna [14]

Combining the terms into one fraction, we have

kx + b - \dfrac{x^3+1}{x^2+1} = \dfrac{(k-1)x^3 + bx^2 + kx + b - 1}{x^2+1}

If this converges to 0 as x\to\infty, then the degree of the numerator must be smaller than the degree of the denominator.

To ensure this, take k=1 and b=0. This eliminates the cubic and quadratic terms in the numerator, and we do have

\displaystyle \lim_{x\to\infty} \frac{x - 1}{x^2 + 1} = \lim_{x\to\infty} \frac{\frac1x - \frac1{x^2}}{1 + \frac1{x^2}} = 0

Alternatively, we can compute the quotient and remainder of the rational expression.

\dfrac{x^3+1}{x^2+1} = x - \dfrac{x-1}{x^2+1}

Then in the limit, we have

\displaystyle \lim_{x\to\infty} \left(kx + b - x + \frac{x-1}{x^2+1}\right) = (k-1) \lim_{x\to\infty} x + b = 0

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3 0
2 years ago
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Nutka1998 [239]

Answer:

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4 years ago
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To the nearest penny, c = $6,252.50

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