<span>h<span>(t)</span>=<span>t<span>34</span></span>−3<span>t<span>14</span></span></span>
Note that the domain of h is <span>[0,∞]</span>.
By differentiating,
<span>h'<span>(t)</span>=<span>34</span><span>t<span>−<span>14</span></span></span>−<span>34</span><span>t<span>−<span>34</span></span></span></span>
by factoring out <span>34</span>,
<span>=<span>34</span><span>(<span>1<span>t<span>14</span></span></span>−<span>1<span>t<span>34</span></span></span>)</span></span>
by finding the common denominator,
<span>=<span>34</span><span><span><span>t<span>12</span></span>−1</span><span>t<span>34</span></span></span>=0</span>
<span>⇒<span>t<span>12</span></span>=1⇒t=1</span>
Since <span>h'<span>(0)</span></span> is undefined, <span>t=0</span> is also a critical number.
Hence, the critical numbers are <span>t=0,1</span>.
I hope that this was helpful.
Answer:
P ( 5 < X < 10 ) = 1
Step-by-step explanation:
Given:-
- Sample size n = 49
- The sample mean u = 8.0 mins
- The sample standard deviation s = 1.3 mins
Find:-
Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.
Solution:-
- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:
X ~ N ( u , s /√n )
Where
s /√n = 1.3 / √49 = 0.2143
- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:
P ( 5 < X < 10 ) = P ( (5 - 8) / 0.2143 < Z < (10-8) / 0.2143 )
= P ( -14.93 < Z < 8.4 )
- Using standard Z-table we have:
P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1
Step-by-step explanation:
Subtracting equation (2) from equation (1)
Substituting d = - 3 in equation (1), we find:
Hence, first term is zero.
