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Jet001 [13]
2 years ago
6

Using the 4 train car strategy for factoring polynomials, what is the green common factor for the 1st two train cars and the gre

en common factor for the 3rd and 4th train car: . note: the answers are intermediary steps. do not enter the final step or factored form, but this question is asking what is the intermediate step as the answer.
2x^2 + 4x + 3xy + 6y

enter your answer in the boxes.
Mathematics
1 answer:
koban [17]2 years ago
7 0

The completely factored expression of 2x^2 + 4x + 3xy + 6y is (2x + 3y)(x + 2)

<h3>How to factor the polynomial?</h3>

The expression is given as:

2x^2 + 4x + 3xy + 6y

Group the expression into two

[2x^2 + 4x] + [3xy + 6y]

Factor out each group

2x(x + 2) + 3y(x + 2)

Factor out x + 2

(2x + 3y)(x + 2)

Hence, the completely factored expression of 2x^2 + 4x + 3xy + 6y is (2x + 3y)(x + 2)

Read more about factored expression at:

brainly.com/question/723406

#SPJ1

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3 years ago
Find the equation of the exponential function represented by the table below
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Answer:

y=4(4)^x

Step-by-step explanation:

Exponential functions are represented in the form y=a(b)^x, where a is the initial value and b is the multiplier.

Here, we see that when x = 0, y = 4. This means that 4 is our initial value, or a. You can look at the a variable almost being like the y-intercept of a graph.

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8 0
3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
irakobra [83]

Answer:

We need a sample size of at least 383.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

85% confidence level

So \alpha = 0.15, z is the value of Z that has a pvalue of 1 - \frac{0.15}{2} = 0.925, so Z = 1.44.

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 85% confidence level with an error of at most 0.03

We need a sample size of at least n.

n is found with M = 0.03, \pi = 0.21

Then

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.44\sqrt{\frac{0.21*0.79}{n}}

0.03\sqrt{n} = 1.44\sqrt{0.21*0.79}

\sqrt{n} = \frac{1.44\sqrt{0.21*0.79}}{0.03}

(\sqrt{n})^{2} = (\frac{1.44\sqrt{0.21*0.79}}{0.03})^{2}

n = 382.23

Rounding up

We need a sample size of at least 383.

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DochEvi [55]
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