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2 years ago

What second degree polynomial function has a leading coefficient of -2 and root 4 with a multiplicity of 2

Mathematics

1 answer:

OverLord2011 [107]2 years ago

8 0

The second degree polynomial with leading coefficient of -2 and root 4 with multiplicity of 2 is:

$p(x) = -2*(x - 4)*(x - 4) = -2*(x - 4)^2$

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

A polynomial of degree N, with the N roots {x₁, ..., xₙ} and a leading coefficient a is written as:

$p(x) = a*(x - x_1)*(x - x_2)*...*(x - x_n)$

Here we know that the degree is 2, the only root is 4 (with a multiplicity of 2, this is equivalent to say that we have two roots at x = 4) and a leading coefficient equal to -2.

Then this polynomial is equal to:

$p(x) = -2*(x - 4)*(x - 4) = -2*(x - 4)^2$

If you want to learn more about polynomials:

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Ty and Gajge are both football players who averaged 4 yards per run in yesterday’s

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Answer:

The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs is the answer.

Step-by-step explanation:

given that Ty and Gajge are football players.

Carries is 15 for both and average is the same 4 for both.

But on scrutiny we find that maximum and minimum and 6 and 2 for Ty.

Hence range for Ty = 6-2 =4 (2 runs on eithre side of mean)

But for Gajge, highest is 19 and lowest is 2.

i.e. range = 19-2 =17 very much higher than that of Ty

The lengths of Gajge’s runs have greater variability because there is a greater difference between his longest and shortest runs.

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3 years ago

DIC A student has a total of $3000 in student loans that will be paid with a 48-month installment loan with monthly payments of

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Step-by-step explanation:

Given that a student has a total of $ 3000 in student loans that will be paid with a 48-month installment loan with monthly payments of $ 73.94, to determine the APR of the loan to the nearest one-half of a percent the following calculation must be done:

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Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Norm

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Answer:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

The distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar x =512$ represent the sample mean

$\sigma=100$ represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ (1)

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

$Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

$25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}$

And we can find the critical value $z_{\alpha/2}$ like this:

$z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576$

And we know that on the right tail of the z score =2.576 we have $\alpha/2$ of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998= $\alpha/2$ , so then $\alpha=0.00498*2=0.009995 \approx 0.01$ , and then we can find the confidence like this:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

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