Hi there!
a. 8x^2 - 1
(x^2 - 4x + 5) + (7x^2 + 4x - 6)
= x^2 + 7x^2 - 4x + 4x - 1
= 8x^2 - 1
b. 5x^2 + 7x - 7
(7x^2 + 4x - 6) - (2x^2 - 3x + 1)
= 7x^2 + 4x - 6 - 2x^2 + 3x - 1
= 7x^2 - 2x^2 + 4x + 3x - 6 - 1
= 5x^2 + 7x -7
Hope this helps!
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. :

Q + d = 100 so q = 100 - d
0.25q + 0.10d = 19
substitute q = 100 - d into 0.25q + 0.10d = 19
0.25q + 0.10d = 19
0.25(100 - d) + 0.10d = 19
25 - 0.25d + 0.10d = 19
-0.15d = -6
d = 40
q = 100 - 40 = 60
answer
Kitara has 60 quarters and 40 dimes
Answer:
Given:
f(x) = 4x² + 1
g(x) = x² - 5
Then
(f + g)(x) = 4x² + 1 + x² - 5
= 5x² - 4
Answer: 5x² - 4