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GarryVolchara [31]
3 years ago
5

I'm not sure how to find zeros and multipilicities

Mathematics
1 answer:
yarga [219]3 years ago
7 0

If p(x) is a polynomial, then the zeros of p(x) are the values of x that satisfy p(x)=0.

So for (1), the zeros of f(x)=(x-4)(x-1)(x+3) are the solutions to

(x-4)(x-1)(x+3)=0\implies x=4,x=1,x=-3

You can think of the multiplicity of each root as the number of times a root satisfies this equation. In this case, the multiplicity of each root is 1, because there is only one factor of x-4 or x-1 or x+3.

For (2), the zeros of g(x) are the solutions to

(x+1)(x-2)^2(x-4)=0\implies x=-1,x=2,x=4

and the multiplicity of the zero x=2 is 2 because there are two factors of x-2 in g(x). The multiplicity of the other roots would be 1.

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Pie
<h2>Answer:</h2>

<h3>container B requires more plastic to make.</h3>

<h2>Step-by-step explanation:</h2>

Container A:

diameter- 7 cm

height- 122 cm

radius- 3.5 cm

<h3>calculating for the surface area</h3>

SA=2×π×r×r+2×π×r×h

SA=2(3.14)(3.5)(3.5)+2(3.14)(3.5)(122)

SA=76.93+2681.56

SA=2758.49 cm^2

<h3>Container B:</h3>

diameter- 11 cm

height- 85 cm

radius- 5.5 cm

<h3>calculating for the surface area</h3>

SA=2×π×r×r+2×π×r×h

SA=2(3.14)(5.5)(5.5)+2(3.14)(5.5)(85)

SA=189.97+2935.90

SA=3125.87 cm^2

<h3>Container B required more plastic to make</h3>
7 0
2 years ago
Pls help good points
liq [111]
I got 3; M=-4 so you plug that in and get -4^2+5(-4)+7
-4 times -4=16
5 times -4=-20
16+(-20)=-4
-4+7=3
8 0
3 years ago
Suppose that E(θˆ1) = E(θˆ2) = θ, V(θˆ 1) = σ2 1 , and V(θˆ2) = σ2 2 . Consider the estimator θˆ 3 = aθˆ 1 + (1 − a)θˆ 2. a Show
katen-ka-za [31]

Answer:

Step-by-step explanation:

Given that:

E( \hat \theta _1) = \theta  \ \ \ \ E( \hat \theta _2) = \theta \ \ \ \ V( \hat \theta _1) = \sigma_1^2  \ \ \ \ V(\hat  \theta_2) = \sigma_2^2

If we are to consider the estimator \hat \theta _3 = a \hat  \theta_1 + (1-a)  \hat \theta_2

a. Then, for  \hat \theta_3 to be an unbiased estimator ; Then:

E ( \hat \theta_3) = E ( a \hat \theta_1+ (1-a) \hat \theta_2)

E ( \hat \theta_3) = aE (  \theta_1) + (1-a) E ( \hat \theta_2)

E ( \hat \theta_3) = a   \theta + (1-a)  \theta = \theta

b) If \hat \theta _1 \ \  and  \ \   \hat \theta_2 are independent

V(\hat \theta _3) = V (a \hat \theta_1+ (1-a) \hat \theta_2)

V(\hat \theta _3) = a ^2 V ( \hat \theta_1) + (1-a)^2 V ( \hat \theta_2)

Thus; in order to minimize the variance of \hat \theta_3 ; then constant a can be determined as :

V( \hat \theta_3) = a^2 \sigma_1^2 + (1-a)^2 \sigma^2_2

Using differentiation:

\dfrac{d}{da}(V \ \hat \theta_3) = 0 \implies 2a \ \sigma_1^2 + 2(1-a)(-1) \sigma_2^2 = 0

⇒

a (\sigma_1^2 + \sigma_2^2) = \sigma^2_2

\hat a = \dfrac{\sigma^2_2}{\sigma^2_1+\sigma^2_2}

This implies that

\dfrac{d}{da}(V \ \hat \theta_3)|_{a = \hat a} = 2 \ \sigma_1^2 + 2 \ \sigma_2^2 > 0

So, V( \hat \theta_3) is minimum when \hat a = \dfrac{\sigma_2^2}{\sigma_1^2+\sigma_2^2}

As such; a = \dfrac{1}{2}       if   \sigma_1^2 \ \ =  \ \  \sigma_2^2

4 0
4 years ago
The rate of change (dP/dt), of the number of people on an ocean beach is modeled by a logistic differential equation. The maximu
Kazeer [188]

Answer:

\frac{dP}{dt} = 2.4P(1 - \frac{P}{1200})

Step-by-step explanation:

The logistic differential equation is as follows:

\frac{dP}{dt} = rP(1 - \frac{P}{K})

In this problem, we have that:

K = 1200, which is the carring capacity of the population, that is, the maximum number of people allowed on the beach.

At 10 A.M., the number of people on the beach is 200 and is increasing at the rate of 400 per hour.

This means that \frac{dP}{dt} = 400 when P = 200. With this, we can find r, that is, the growth rate,

So

\frac{dP}{dt} = rP(1 - \frac{P}{K})

400 = 200r(1 - \frac{200}{1200})

166.67r = 400

r = 2.4

So the differential equation is:

\frac{dP}{dt} = rP(1 - \frac{P}{K})

\frac{dP}{dt} = 2.4P(1 - \frac{P}{1200})

3 0
3 years ago
Show on the graph how much honey is needed for 60 grams of salt
inessss [21]

Answer:

there is no pic

Step-by-step explanation:

6 0
3 years ago
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