50p thanks to whoever helps :)
1 answer:
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Integrate both sides with respect to <em>t</em> :
∫ d<em>y</em>/d<em>t</em> d<em>t</em> = ∫ -12<em>t</em> ² d<em>t</em>
<em>y(t)</em> = -4<em>t</em> ³ + <em>C</em>
Use the initial condition to solve for <em>C</em> :
5 = -4•0³+ <em>C</em>
<em>C</em> = 5
So
<em>y(t)</em> = -4<em>t</em> ³ + 5
and the answer is D.
Alternatively, you can directly apply the fundamental theorem of calculus:
Answer:
A task time of 177.125s qualify individuals for such training.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by
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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem, we have that:
A distribution that can be approximated by a normal distribution with a mean value of 145 sec and a standard deviation of 25 sec, so .
The fastest 10% are to be given advanced training. What task times qualify individuals for such training?
This is the value of X when Z has a pvalue of 0.90.
Z has a pvalue of 0.90 when it is between 1.28 and 1.29. So we want to find X when .
So
A task time of 177.125s qualify individuals for such training.