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antoniya [11.8K]

3 years ago

3" alt=" - y = \frac{2}{5}x - 3" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

raketka [301]3 years ago

5 0

The first step to solving this equation is to change the signs on both sides of the equation
y = - $\frac{2}{5}$ x + 3
Since we don't know what x is but we do know why y is,, we will write our answer with the answer to y and x being left as undefined.
y = - $\frac{2}{5}$ x + 3, x ∈ R
Let me know if you have any further questions
:)

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What does A B and C =?

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Answer:

z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

Step-by-step explanation:

We must evaluate the differences of the means of the two machines, to do so, we will assume a CI of 95%, and as the interest is to find out if the new machine has better performance ( machine has a bigger efficiency or the new machine produces more units per unit of time than the old one) the test will be a one tail-test (to the left).

New machine

Sample mean x₁ = 25

Sample variance s₁ = 27

Sample size n₁ = 45

Old machine

Sample mean x₂ = 23

Sample variance s₂ = 7,56

Sample size n₂ = 36

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Null hypothesis H₀ x₂ - x₁ = d = 0

Alternative hypothesis Hₐ x₂ - x₁ < 0

CI = 90 % ⇒ α = 10 % α = 0,1 z(c) = - 1,28

To calculate z(s)

z(s) = ( x₂ - x₁ ) / √s₁² / n₁ + s₂² / n₂

s₁ = 27 ⇒ s₁² = 729

n₁ = 45 ⇒ s₁² / n₁ = 16,2

s₂ = 7,56 ⇒ s₂² = 57,15

n₂ = 36 ⇒ s₂² / n₂ = 1,5876

√s₁² / n₁ + s₂² / n₂ = √ 16,2 + 1.5876 = 4,2175

z(s) = (23 - 25 )/4,2175

z(s) = - 0,4742

Comparing z(s) and z(c)

|z(s)| < | z(c)|

z(s) is in the acceptance region. We accept H₀ we did not find a significantly difference in the performance of the two machines therefore we suggest not to buy a new machine

The very hight dispersion of values s₁ = 27 is evidence of frecuent values quite far from the mean

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3 years ago

Which polynomials are prime? check all that apply. 15x2 10x – 9x 7 20x2 – 12x 30x – 18 6x3 14x2 – 12x – 28 8x3 20x2 3x 12 11x4 4

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The options that are prime polynomials are,

15x^2 +10x -9x+7, 8x^3+20x^2+3x+12 and 11x^4+4x^2-6x^2-16

<h3>What is the prime polynomial?</h3>

A polynomial with integer coefficients that cannot be factored into polynomials of lower degree, also with integer coefficients, is called an irreducible or prime polynomial.

To determine all options in order to check that the polynomials are prime or not:

1). 15x^2 +10x -9x+7

5x(3x + 2) - (9x - 7)

Therefore, this polynomial can not be factored into lower degree polynomial. Therefore, it is a prime polynomial.

2). 20x^2-12x+30x-18

4x(5x - 3)+6(5x-3)

(4x + 6)(5x - 3)

Therefore, this polynomial is converted into a lower degree polynomial. Therefore, it is not a prime polynomial.

3). 6x^3+14x^2-12x-28

2x^2(3x+7)-4(3x+7)

(2x^2-4)(3x+7)

Therefore, this polynomial is converted into a lower degree polynomial. Therefore, it is not a prime polynomial.

4). 8x^3+20x^2+3x+12

4x^2(2x+5)+(3x+20)

Therefore, this polynomial can not be factored into lower degree polynomial. Therefore, it is a prime polynomial.

5). 11x^4+4x^2-6x^2-16

x^2(11x^2+4)-2(3x^2+8)

So, this polynomial can not be factored into lower degree polynomial. Therefore, it is a prime polynomial.

To learn more about the prime polynomial visit:

brainly.com/question/16749843

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