Answer:
The minimum level for which the battery pack will be classified as highly sought-after class is 2.42 hours
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the minimum level for which the battery pack will be classified as highly sought-after class
At least the 100-10 = 90th percentile, which is the value of X when Z has a pvalue of 0.9. So it is X when Z = 1.28.




The minimum level for which the battery pack will be classified as highly sought-after class is 2.42 hours
69.24 because you round the three up
Answer:
Step-by-step explanation:
If you're looking for what the half angle of the tangent of theta is, I'm a bit confused as to why you think the angle in the 4th quadrant, x, is relevant. But maybe you don't know it isn't and it's a "trick" to throw you off. Hmm...
Anyways, the half angle identity for tangent is

There are actually 3 identities for the tangent of a half angle, but this one works just as well as either of the others do, so I'm going with this one.
If theta is in QIII, the value of -4 goes along the x axis and the hypotenuse is 5. That makes the missing side, by Pythagorean's Theorem, -3. Filling in our formula:
which simplifies a bit to
and a bit more to

Bring up the lower fraction and flip it to divide to get
which of course simplifies to
-3. Choice A.
Step-by-step explanation:
<em>
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<em>This </em><em>is </em><em>the</em><em> </em><em>original </em><em>equation</em><em> </em><em>so</em><em> </em><em>you </em><em>will</em><em> </em><em>open</em><em> </em><em>the</em><em> </em><em>brackets</em><em>.</em>
<em>
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<em>The</em><em> </em><em>x</em><em> </em><em>will</em><em> </em><em>cross</em><em> </em><em>the</em><em> </em><em>division </em><em>sign</em><em> </em><em>making</em><em> </em><em>it</em><em> </em><em>times</em><em> </em><em>x</em><em>.</em>
<em>
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<em>So</em><em> </em><em>you </em><em>will</em><em> </em><em>proceed</em><em> </em><em>to</em><em> </em><em>multiply</em><em> </em><em>the</em><em> </em><em>9</em><em>x</em><em> </em><em>by</em><em> </em><em>x</em><em> </em><em>making </em><em>it</em><em>;</em>
<em>
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<em>So</em><em> </em><em>you</em><em> </em><em>will</em><em> </em><em>subtract</em><em> </em><em>9</em><em>x</em><em>²</em><em> </em><em>from</em><em> </em><em>3</em><em>x</em><em>²</em><em> </em><em>which</em><em> </em><em>won't </em><em>be</em><em> </em><em>possible </em><em>so</em><em> </em><em>it </em><em>will</em><em> </em><em>be</em><em> </em><em>negative</em><em>.</em>
<em>
</em>
<em>Then</em><em> </em><em>you </em><em>will</em><em> </em><em>square</em><em> </em><em>root</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>making</em><em> </em><em>it</em><em>;</em>
<em>
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<em>So</em><em> </em><em>the</em><em> </em><em>²</em><em> </em><em>will</em><em> </em><em>cancel</em><em> </em><em>the</em><em> </em><em>√</em><em> </em><em>leaving </em>
<em>-6x</em><em>=</em><em>√</em><em>3</em>
<em>-6x</em><em>=</em><em>1</em><em>.</em><em>7</em><em>3</em><em>(</em><em>Divide</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>by</em><em> </em><em>-</em><em>6</em><em>)</em>
<em>x</em><em>=</em><em>0</em><em>.</em><em>2</em><em>8</em><em>8</em>
<em>That</em><em>'s</em><em> </em><em>the</em><em> </em><em>final </em><em>answer</em><em>.</em>