M(x) = 4x^3 - 5x^2 - 7x
Let us first find the zeros of the function.
That is when it is equal to zero.
m(x) = 4x^3 - 5x^2 - 7x = 0
x(4x^2 - 5x - 7) = 0. Therefore x = 0 or 4x^2 - 5x - 7 = 0.
Using a quadratic function calculator to solve 4x^2 - 5x - 7
x = 2.09, -0.84
Therefore the zeros are x =-0.84, 0, 2.09 for the function m(x).
The intervals observed are imagining that the zeros are on the number line:
x<-0.84, -0.84<x<0, 0<x<2.09, x>2.09.
For each of this range we would test the function with a number that falls in the range.
The function is decreasing in the interval where it is less than 0.
For x<-0.84, let us test x = -1, m(x) = 4x^3 - 5x^2 - 7x = 4(-1)^3 - 5(-1)^2 - 7(-1) = -4 -5 +7 = -2, -2 < 0, so it is decreasing here.
For -0.84<x<0, let us test x = -0.5, m(x) = 4x^3 - 5x^2 - 7x = 4(-0.5)^3 - 5(-0.5)^2 - 7(-0.5) = -0.5 -1.25 +3.5 = 1.75, 1.75 >0. It is not decreasing.
For 0<x<2.09, let us test x = 1, m(x) = 4x^3 - 5x^2 - 7x =
4(1)^3 - 5(1)^2 - 7(1) = 4 -5 -7 = -8, -8 <0. It is decreasing.
For x>2.09, let us test x = 3, m(x) = 4x^3 - 5x^2 - 7x =
4(3)^3 - 5(3)^2 - 7(3) = 108 -45 -21 = 42, 42 >0. It is not decreasing.
So the function is decreasing in the intervals:
x < -0.84, & 0<x<2.09.
Answer:
a) Binomial.
b) n=20, p=0.01, k≥2
The probability hat a package sold will be refunded is P=0.0169.
Step-by-step explanation:
a) We know that
- the defective probability is constant and independent.
- the sample size is bigger than one subject.
The most appropiate distribution to represent this random variable is the binomial.
b) The parameters are:
- Sample size (amount of clips in the package): n=20
- Probability of defective clips: p=0.01.
- number of defective clips that trigger the money-back guarantee: k≥2
The probability of the package being refunded can be calculated as:
Answer:
The angles are vertical angles and they are congruent
Step-by-step explanation:
The answer for this problem is 1.7 I think I’m not sure
Answer: 
b) 5314
c) ln 2.7
d) 4.6 hrs
<u>Step-by-step explanation:</u>
