How do you solve this??? I need it fast please help with steps. (x+3)(x+1)^2(x−4)>0
1 answer:
Answer:
x < -3 or x > 4
Step-by-step explanation:
The product of the factors is 4th-degree, and the leading coefficient is 1 (positive).
The factors tell you the following about the zeros:
x = -3 . . . sign change from + to -
x = -1 . . . graph touches, but no sign change. - on either side of x=-1
x = 4 . . . sign change from - to +
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The function will be positive for x < -3 and for x > +4.
_____
<em>Additional comments</em>
The value of the function is zero when any of its factors is zero. The value of a factor changes sign when the value of x changes from one side of the 0 to the other. For example, here, we have x+1 as a factor, so x=-1 is a zero. When x = -0.9, the factor is positive (-0.9 +1 = 0.1). When x = -1.1, the factor is negative (-1.1 +1 = -0.1). If this factor were to the first power, it would cause the value of the function to change sign at x=-1.
However, the factor (x+1) has an even power: (x+1)^2. That means when the factor is negative, its square is positive, and when the factor is positive, its square is positive. In short, the factor (x+1)^2 causes the function to be zero at x=-1, but does not make the function change sign there.
The product of all of these factors will result in a polynomial with x^4 as the highest-degree term. That means the function is of even degree. The leading coefficient is 1, so is positive, and the function will generally have a U-shape. The left-most (odd-degree) zero will be where the function changes sign from positive to negative. The right-most (odd-degree) zero will be where the function changes sign from negative to positive. Those zeros are x=-3 and x=+4, respectively. There are no places between these where the function changes sign, so these zeros are the ends of the regions where the function is positive (>0).
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Answer:
the probability that we hit the bullseye at least 100 times is 0.0113
Step-by-step explanation:
Given the data in the question;
Binomial distribution
We find the probability of hitting the dart on the disk
⇒ Area of small disk / Area of bigger disk
⇒ πR₁² / πR₂²
given that; disk-shaped board of radius R² = 5, disk-shaped bullseye with radius R₁ = 1
so we substitute
⇒ π(1)² / π(5)² = π/π25 = 1/25 = 0.04
Since we have to hit the disk 2000 times, we represent the number of times the smaller disk ( BULLSEYE ) will be hit by X.
so
X ~ Bin( 2000, 0.04 )
n = 2000
p = 0.04
np = 2000 × 0.04 = 80
Using central limit theorem;
X ~ N( np, np( 1 - p ) )
we substitute
X ~ N( 80, 80( 1 - 0.04 ) )
X ~ N( 80, 80( 0.96 ) )
X ~ N( 80, 76.8 )
So, the probability that we hit the bullseye at least 100 times, P( X ≥ 100 ) will be;
we covert to standard normal variable
⇒ P( X ≥ )
⇒ P( X ≥ 2.28217 )
From standard normal distribution table
P( X ≥ 2.28217 ) = 0.0113
Therefore, the probability that we hit the bullseye at least 100 times is 0.0113