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vredina [299]
2 years ago
9

Nswer two questions about Systems A AA and B BB: System A AA start text, end text System B BB { − 3 x + 12 y = 15 7 x − 10 y = −

2 ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ​ −3x+12y=15 7x−10y=−2 ​ { − x + 4 y = 5 7 x − 10 y = − 2 ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ ​ −x+4y=5 7x−10y=−2 ​ 1) How can we get System B BB from System A AA? Choose 1 answer: Choose 1 answer: (Choice A) A Replace one equation with the sum/difference of both equations (Choice B) B Replace only the left-hand side of one equation with the sum/difference of the left-hand sides of both equations (Choice C) C Replace one equation with a multiple of itself (Choice D) D Replace one equation with a multiple of the other equation
Mathematics
1 answer:
rjkz [21]2 years ago
7 0

Answer:

(Choice C) C Replace one equation with a multiple of itself

Step-by-step explanation:

Since system A has the equations

-3x + 12y = 15 and 7x - 10y = -2 and,

system B has the equations

-x + 4y = 5 and 7x - 10 y = -2.

To get system B from system A, we notice that equation -x + 4y = 5 is a multiple of -3x + 12y = 15 ⇒ 3(-x + 4y = 5) = (-3x + 12y = 15).

So, (-x + 4y = 5) = (1/3) × (-3x + 12y = 15)

So, we replace the first equation in system B by 1/3 the first equation in system A to obtain the first equation in system B.

So, choice C is the answer.

We replace one equation with a multiple of itself.

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Answer:

<em>c=6, d=2</em>

Step-by-step explanation:

<em>Equations </em>

We must find the values of c and d that make the below equation be true

\sqrt[3]{162x^cy^5}=3x^2y \sqrt[3]{6y^d}

Let's cube both sides of the equation:

\left (\sqrt[3]{162x^cy^5}\right )^3=\left (3x^2y \sqrt[3]{6y^d}\right)^3

The left side just simplifies the cubic root with the cube:

162x^cy^5=\left (3x^2y \sqrt[3]{6y^d}\right)^3

On the right side, we'll simplify the cubic root where possible and power what's outside of the root:

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Simplifying

x^cy^5=x^6y^{3+d}

Equating the powers of x and y separately we find

c=6

5=3+d

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The values are

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2 years ago
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The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deduction
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Answer:

(a) <em>                             </em><em>n</em> :      20           50          100         500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(b) The correct option is (b).

Step-by-step explanation:

Let the random variable <em>X</em> represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.

The mean amount of deductions is, <em>μ</em> = $16,642 and standard deviation is, <em>σ</em> = $2,400.

Assuming that the random variable <em>X </em>follows a normal distribution.

(a)

Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:

  • For a sample size of <em>n</em> = 20

P(\mu-200

                                           =P(-0.37

  • For a sample size of <em>n</em> = 50

P(\mu-200

                                           =P(-0.59

  • For a sample size of <em>n</em> = 100

P(\mu-200

                                           =P(-0.83

  • For a sample size of <em>n</em> = 500

P(\mu-200

                                           =P(-1.86

<em>                                  n</em> :      20           50          100         500

P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886    0.4444    0.5954    0.9376

(b)

The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample (\bar x) approaches the whole population mean (\mu_{x}).

Consider the probabilities computed in part (a).

As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.

So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.

Thus, the correct option is (b).

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This question is incomplete because it lacks the appropriate options

Complete Question

Fifteen percent of the students at Central High take Spanish, 12% take French, and 4% take both. If a student is selected at random, what is the

probability that the student takes Spanish or French?

A 31%

B. 27%

C. 23%

D. 19%

Answer:

C. 23%

Step-by-step explanation:

At Central High

From the question, we are told that:

Spanish = 15%

French = 12%

Spanish and French = 4%

French or Spanish = unknown

Therefore, the probability that a student at random takes Spanish or French

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= 15% + 12% - 4%

= 27% - 4%

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