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2 years ago

(6-2i)^2 which is the coefficient of i ?

Mathematics

1 answer:

Yuki888 [10]2 years ago

7 0

Option A: -24 is the coefficient of i

Explanation:

The expression is $(6-2 i)^{2}$

To determine the coefficient of i, first we shall find the square of the binomial for the expression $(6-2 i)^{2}$

The formula to find the square of the binomial for this expression is given by

$(a-b)^{2}=a^{2}-2 a b+b^{2}$

where $a=6$ and $b=2i$

Substituting this value and expanding, we get,

$(6-2 i)^{2}=6^{2} -2(6)(2i)+(2i)^{2}$

Simplifying the terms, we have,

$(6-2 i)^{2}=36-24i-4$

Thus, from the above expression the coefficient of i is determined as -24.

Hence, Option A is the correct answer.

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2 years ago

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MA_775_DIABLO [31]

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Step-by-step explanation:

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2 years ago

The speeds of vehicles traveling on a highway are normally distributed with an unkown population mean and standard deviation. A

Maksim231197 [3]

Answer:

The 90% confidence interval would be given by (60.09;69.91)

We are 90% confident that the true mean for the speeds of vehicles traveling on a highway is between 60.09 and 69.91 miles per hour.

Step-by-step explanation:

1) 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".

$\bar X =65$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=9 represent the sample standard deviation

n=11 represent the sample size

2) Confidence interval

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:

$df=n-1=11-1=10$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,10)".And we see that $t_{\alpha/2}=1.81$

Now we have everything in order to replace into formula (1):

$65-1.81\frac{9}{\sqrt{11}}=60.09$

$65+1.81\frac{9}{\sqrt{11}}=69.91$

So on this case the 90% confidence interval would be given by (60.09;69.91)

We are 90% confident that the true mean for the speeds of vehicles traveling on a highway is between 60.09 and 69.91 miles per hour.

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2 years ago