Answer: a. 0.6759 b. 0.3752 c. 0.1480
Step-by-step explanation:
Given : The long-distance calls made by the employees of a company are normally distributed with a mean of 6.3 minutes and a standard deviation of 2.2 minutes
i.e. 
minutes
minutes
Let x be the long-distance call length.
a. The probability that a call lasts between 5 and 10 minutes will be :-
b. The probability that a call lasts more than 7 minutes. :
![P(X>7)=P(\dfrac{X-\mu}{\sigma}>\dfrac{7-6.3}{2.2})\\\\=P(Z>0.318)\ \ \ \ [z=\dfrac{X-\mu}{\sigma}]\\\\=1-P(Z](https://tex.z-dn.net/?f=P%28X%3E7%29%3DP%28%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%3E%5Cdfrac%7B7-6.3%7D%7B2.2%7D%29%5C%5C%5C%5C%3DP%28Z%3E0.318%29%5C%20%5C%20%5C%20%5C%20%5Bz%3D%5Cdfrac%7BX-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-P%28Z%3C0.318%29%5C%5C%5C%5C%3D1-0.6248%5C%20%5C%20%5C%20%5C%20%5B%5Ctext%7Bby%20z-table%7D%5D%5C%5C%5C%5C%3D0.3752)
c. The probability that a call lasts more than 4 minutes. :
Answer:
Step-by-step explanation:
Answer:
Any [a,b] that does NOT include the x-value 3 in it.
Either an [a,b] entirely to the left of 3, or
an [a,b] entirely to the right of 3
Step-by-step explanation:
The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.
Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.
Answer:
x= −297
/13
Step-by-step explanation:
