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iVinArrow [24]
3 years ago
12

A digital scale reports a 10kg weight as weighing 8.975kg. which of the following is true?

Mathematics
1 answer:
lukranit [14]3 years ago
6 0
That question is accompanied by these answer choices:

<span>A. The scale is accurate but not precise.
B. The scale is precise but not accurate.
C. The scale is neither precise nor accurate.
D. The scale is both accurate and precise.

Then you need to distinguish between accuracy and precision.

Accuracy refers to the closeness of the measure to the real value, while precision, in this case, refers to the level of significant figures that the sacle report.

The fact that the scale reports the number with 4 significant figures means that it is very precise, but the fact that the result is not so close to the real value as the number of significan figures pretend to be, means that the scale is not accurate.

So, the answer is that the scale is precise but not accurate (the option B).
</span>
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yes.<em> </em><em>2</em><em> </em><em>is</em><em> </em><em>an</em><em> </em><em>only</em><em> </em><em>even</em><em> </em><em>prime</em><em> </em><em>number</em><em>. </em>

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<em>2</em><em> </em><em>is</em><em> </em><em>divisible</em><em> </em><em>by</em><em> </em><em>1</em><em> </em><em>and</em><em> </em><em>2</em>

<em>prime</em><em> </em><em>number</em><em> </em><em>has</em><em> </em><em>only</em><em> </em><em>two</em><em> </em><em>factors</em><em>. </em>

<em>FOR</em><em> </em><em>Example</em><em>:</em><em> </em><em>13</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

6 0
3 years ago
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8 0
3 years ago
30 Points please help!! Please explain too!
sergiy2304 [10]
Initial: 200 feet = opp. side
Angle = 17.31, adj. Side = a
So tan (17.31) = opp. Side / adj. Side
tan (17.31) = 200 / a
--> a = 200/tan17.31 = 641.73
When stopped: adj. Still = a
Opp. Side still = 200, angle = 46.41
tan46.41 = 200 / a
--> a = 200/tan46.41 = 190.39
Now subtract those two distances to get the distance traveled from initial to when it stopped:
641.73 - 190.39 = 451.34 feet
3 0
3 years ago
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