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1 year ago

Which equivalent four-term polynomial can be created using the X method?

Mathematics

2 answers:

docker41 [41]1 year ago

8 0

Answer:

C.x2 + 12x + 4x + 48

Step-by-step explanation:

Dmitry [639]1 year ago

7 0

Answer:

The answer is option C. ( $x^{2}$ + 12x + 4x + 48)

Step-by-step explanation:

12x+4x

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A random sample of 50 units is drawn from a production process every half hour. the true fraction of nonconforming products manu

Naya [18.7K]

solution:

The probability mass function for binomial distribution is,

Where,

X=0,1,2,3,…..; q=1-p

find the probability that (p∧ ≤ 0.06) , substitute the values of sample units (n) , and the probability of nonconformities (p) in the probability mass function of binomial distribution.

Consider x to be the number of non-conformities. It follows a binomial distribution with n being 50 and p being 0.03. That is,

binomial (50,0.02)

Also, the estimate of the true probability is,

p∧ = x/50

The probability mass function for binomial distribution is,

Where,

X=0,1,2,3,…..; q=1-p

The calculation is obtained as

P(p^ ≤ 0.06) = p(x/20 ≤ 0.06)

= 50cx ₓ (0.03)x ₓ (1-0.03)50-x

= (50c0 ₓ (0.03)0 ₓ (1-0.03)50-0 + 50c1(0.03)1 ₓ (1-0.03)50-1 + 50c2 ₓ (0.03)2 ₓ (1-0.03)50-2 +50c3 ₓ (0.03)3 ₓ (1- 0.03)50-3 )

=( ₓ (0.03)0 ₓ (1-0.03)50-0 + ₓ (0.03)1 ₓ (1-0.03)50-1 + ₓ (0.03)2 ₓ (1-0.03)50-2 ₓ (0.03)3 ₓ (1-0.03)50-3 )

5 0

2 years ago

At the time of her grandson's birth, a grandmother deposits $ 2000 in an account that pays 3.5 % compounded monthly. What will

11Alexandr11 [23.1K]

At the time of her grandson's birth, a grandmother deposits $12,000.00 in an account that pays 2% compound monthly. What will be that value of the account at the child's twenty-first birthday, assuming that no other deposits or withdrawls are made during the period.
---
A(t) = P(1+(r/n))^(nt)
---
A(21) = 12000(1+(0.02/12))^(12*21)
---
A(21) = 12000(1.5214)
---
A(21) = #18,257.15

7 0

2 years ago

Read 2 more answers

The mean score of a competency test is 82, with a standard deviation of 2. Between what two values do about 99.7% of the values

goldenfox [79]

99.7% encompasses about 3 standard deviations either side of the mean.

82 ±3*2 = (76, 88)

About 99.7% of the values lie between 76 and 88.

5 0

1 year ago

Read 2 more answers

Find the quantity q that maximizes profit if the total revenue, R(q), and total cost, C(q) are given in dollars by R(q)=3q−0.002

marysya [2.9K]

Answer:

The value of q that maximize the profit is q=200 units

Step-by-step explanation:

we know that

The profit is equal to the revenue minus the cost

we have

$R(q)=3q-0.002q^{2}$ ---> the revenue

$C(q)=200+2.2q$ ---> the cost

The profit P(q) is equal to

$P(q)=R(q)-C(q)$

substitute the given values

$P(q)=(3q-0.002q^{2})-(200+2.2q)$

$P(q)=3q-0.002q^{2}-200-2.2q$

$P(q)=-0.002q^{2}+0.8q-200$

This is a vertical parabola open downward (because the leading coefficient is negative)

The vertex represent a maximum

The x-coordinate of the vertex represent the value of q that maximize the profit

The y-coordinate of the vertex represent the maximum profit

using a graphing tool

Graph the quadratic equation

The vertex is the point (200,-120)

see the attached figure

therefore

The value of q that maximize the profit is q=200 units

6 0

2 years ago

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |

Bas_tet [7]

Answer:

$M=168k$

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

Step-by-step explanation:

Let's begin with the mass definition in terms of density.

$M=\int\int \rho dA$

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

$M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx$

Let's solve this integral:

$M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx$

$M=k\int^{9}_{1}21dx$

$M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}$

So the mass will be:

$M=21k*8=168k$

Now we need to find the x-coordinate of the center of mass.

$\bar{x}=\frac{1}{M}\int\int x*\rho dydx$

$\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx$

$\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx$

$\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx$

$\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}$

$\bar{x}=\frac{21}{168}*40=5$

Now we need to find the y-coordinate of the center of mass.

$\bar{y}=\frac{1}{M}\int\int y*\rho dydx$

$\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx$

$\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx$

$\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx$

$\bar{y}=\frac{255}{672}\int^{9}_{1}dx$

$\bar{y}=\frac{255}{672}8=\frac{2040}{672}$

$\bar{y}=\frac{85}{28}$

Therefore the center of mass is:

$(\bar{x},\bar{y})=(5,\frac{85}{28})$

I hope it helps you!

3 0

2 years ago