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3 years ago

Answer this question

Mathematics

1 answer:

Murljashka [212]3 years ago

5 0

Answer:

a = p * q

b = p * s + q * r

c = r * s

Step-by-step explanation:

In the trinomial ax² + bx + c

a is the coefficient of x²

b is the coefficient of x

c is the numerical term

∵ The trinomial is ax² + bx + c

∵ Its factors are (px + r) and (qx + s)

∴ ax² + bx + c = (px + r)(qx + s)

∵ (px + r)(qx + s) = (px)(qx) + (px)(s) + r(qx) + (r)(s)

∴ (px + r)(qx + s) = pqx² + (psx + qrx) + rs

∴ ax² + bx + c = pqx² + (ps + qr)x + rs

→ By comparing the two sides

∵ ax² = pqx² ⇒ divide both sides by x²

∴ a = pq

∵ bx = (ps + qr)x ⇒ Divide both sides by x

∴ b = ps + qr

∴ c = rs

∴ a = p * q

∴ b = p * s + q * r

∴ c = r * s

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A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec

VMariaS [17]

Answer:

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

$\bar X=19.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=9-1=8$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that $t_{\alpha/2}=$