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:
-1
Step-by-step explanation:
Plug in given values
-4x + 5y + 6z
-4(-2) + 5(3) + 6(-4)
8 + 15 -24
23-24
-1
Probably b. flyer
Of course, it's tough to tell without the selection.
Answer:
112 if you don't include the triangle sides as they are not necessary, but 124 if you include one triangle and 136 if you include both triangle sides
Answer:
14.2cm
Step-by-step explanation:
Complete question:
<em>In circle O, the length of radius OL is 6 cm and the length of arc LM is 6.3 cm. The measure of angle MON is 75°. </em>
<em>Rounded to the nearest tenth of a centimeter, what is the length of arc LMN? </em>
<em></em>
Find the diagram attached
arc LN= arc LM + arc MN
First we need to get the arc MN
length of an arc = theta/360 * 2πr
length of arc MN = 75/360 * 2(3.14)(6)
length of arc MN = 0.20833*37.68
length of arc MN = 7.85
Hence;
arc LN = 6.3 + 7.85
arc LN = 14.15cm
Hence the length of arc LN is 14.2cm