Answer:
thank you so much for free points
and I don't know the answer but really sorry I don't want to ask some questions on this that's why I did this and thank you for free points
![y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\ y''(0)=20\cdot0^3=0](https://tex.z-dn.net/?f=y%3Dx%5E5-3%5C%5C%20y%27%3D5x%5E4%5C%5C%5C%5C%205x%5E4%3D0%5C%5C%20x%3D0%5C%5C%200%5Cin%20%5B-2%2C1%5D%5C%5C%5C%5C%20y%27%27%3D20x%5E3%5C%5C%5C%5C%0Ay%27%27%280%29%3D20%5Ccdot0%5E3%3D0)
The value of the second derivative for

is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of

is always positive for

. That means at

there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval
![[-2,1]](https://tex.z-dn.net/?f=%5B-2%2C1%5D)
.
The function

is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.
Answer:
-2, C
Step-by-step explanation:
Use the formula: (y2-y1)/(x2-x1)
Substitute the numbers:
(4-(-4)) / (3-7)
=8 / -4
= -2
The slope is -2, or C
No.
0.3^4 or 0.3x0.3x0.3x0.3= 0.0081
0.9^2 or 0.9x0.9=0.81
These are not equivalent.
Answer:
The Answer is 76.
Step-by-step explanation:
Given the normal distribution " 10% of employees (rated) exemplary, 20% distinguished, 40% competent, 20% marginal, and 10% unacceptable'', we can see that exemplary employees are top 10% rated employees.
We have the formula for normal distribution:
z=(X-M)÷σ
where z is the <em>minimum z-score </em>for top 10% employee, X is the <em>minimum </em>score for top 10% employee, M is the <em>mean</em> of the score distribution, σ is the <em>standard deviation</em> of the score distribution.
The z-score we are looking for is the value "a" that separates the highest 10% from the lowest 90% i.e. P(z≤a)=0.90
If we look at z-table, corresponding value for a is 1.28155
We can now put the values in the formula:
1.28155=
So X=(1.28155×20)+50=75.631
Therefore minimum score for exemplary employee is 76.