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3 years ago

Choose all the polynomials that are in standard form?

Mathematics

2 answers:

riadik2000 [5.3K]3 years ago

8 0

The answer should be B

Allisa [31]3 years ago

6 0

<h3>2 Answers: Choice B and Choice D</h3>

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Explanation:

Standard form will always have the largest exponents listed first on the left side. Then as you move to the right, the exponents will decrease. Choice B shows this with the exponents counting down (3,2). Choice D is a similar story with the exponents counting down (9,2,1,0). Think of -6x as -6x^1, and also think of the 10 as 10x^0.

Something like choice A is a non-answer because the term with the largest exponent 3 is buried in the middle, and not at the left side. The exponents are not in decreasing order. Choice C can be ruled out for similar reasons.

Side note: the largest exponent is the degree of the polynomial. This only applies to single variable polynomials.

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Answer:

m<A=94

Step-by-step explanation:

(2x+10)+(2x+2)=180

4x+12=180

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4x=168

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x=42

m<A= 2x+10

m<A=2(42)+10

m<A=84+10

m<A=94

Check:

Add m<A and m<B which should equal 180.

m<B=2x+2

m<B=2(42)+2

m<B=84+2

m<B=86

94+86=180

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The 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]

<h3>
How to find the confidence interval for population mean from large samples (sample size > 30)?</h3>

Suppose that we have:

Sample size n > 30
Sample mean = $\overline{x}$
Sample standard deviation = s
Population standard deviation = $\sigma$
Level of significance = $\alpha$

Then the confidence interval is obtained as

Case 1: Population standard deviation is known

$\overline{x} \pm Z_{\alpha /2}\dfrac{\sigma}{\sqrt{n}}$

Case 2: Population standard deviation is unknown.

$\overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}$

For this case, we're given that:

Sample size n = 90 > 30
Sample mean = $\overline{x}$ = 138
Sample standard deviation = s = 34
Level of significance = $\alpha$ = 100% - confidence = 100% - 90% = 10% = 0.1 (converted percent to decimal).

At this level of significance, the critical value of Z is: $Z_{0.1/2}$ = ±1.645

Thus, we get:

$CI = \overline{x} \pm Z_{\alpha /2}\dfrac{s}{\sqrt{n}}\\CI = 138 \pm 1.645\times \dfrac{34}{\sqrt{90}}\\\\CI \approx 138 \pm 5.896\\CI \approx [138 - 5.896, 138 + 5.896]\\CI \approx [132.104, 143.896] \approx [130.10, 143.90]$

Thus, the 90% confidence interval for the population mean of the considered population from the given sample data is given by: Option C: [130.10, 143.90]

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