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3 years ago

The lines y=3x-1 and y=ax+2 are perpendicular if a=___

Mathematics

2 answers:

mezya [45]3 years ago

8 0

Answer:

$a = \frac{-1}{3}.$

Step-by-step explanation:

The equation of line in the form .

y = mx + c

Where m is the slope and c is the y- intercept .

As given

The lines y=3x-1 and y=ax+2 are perpendicular .

Here 3 is slope for equation of line y=3x-1 and a is slope for equation of line

y=ax+2 .

Now by using properties of the perpendicular lines property .

When two lines are perpendicular than slope of one line is negative reciprocal of the other line .

Thus

$a = \frac{-1}{3}$

Therefore $a = \frac{-1}{3}.$

givi [52]3 years ago

4 0

To find the perpendicular slope, you find the reciprocal of the slope (in this case, 1/3) and multiply that by -1 (in this case, it gets -1/3)

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Semmy [17]

Exact Area = pi square units

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Work Shown:

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3 years ago

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RSB [31]

Answer:

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Step-by-step explanation:

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3 years ago

The TreeDropp’d Fruit Company wants to sell its apples overseas in attractive pink and yellow gift boxes. To design the boxes, t

SashulF [63]

Answer:

a)For this case the best estimator for the true mean is the sample mean $\hat \mu =\bar X$ because:

$E(\bar X)= \frac{\sum_{i=1}^n E(X_i)}{n}$

And if we assume that each observation $X_1 , X_2,...,X_n$ follows a normal distribution $X_i \sim N(\mu,\sigma)$ then we have:

$E(\bar X)=\frac{1}{n} n\mu =\mu$

So then yes the $\bar X[/tex[ is a unbiased estimator for the true mean. b) [tex]z_{\alpha/2}=1.96$

c) $ME= 1.96 \frac{0.4}{\sqrt{50}}=0.1109$

Step-by-step explanation:

Previous concepts

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".

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".

$\bar X=4.1$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=0.4$ represent the population standard deviation

n=50 represent the sample size

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma=0.4$

The sample mean $\bar X$ is distributed on this way:

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

A. What is the point estimate in this scenario? Is it an unbiased estimator?

For this case the best estimator for the true mean is the sample mean $\hat \mu =\bar X$ because:

$E(\bar X)= \frac{\sum_{i=1}^n E(X_i)}{n}$

And if we assume that each observation $X_1 , X_2,...,X_n$ follows a normal distribution $X_i \sim N(\mu,\sigma)$ then we have:

$E(\bar X)=\frac{1}{n} n\mu =\mu$

So then yes the $\bar X[/tex[ is a unbiased estimator for the true mean. B. What is the critical z-value for a 95% interval?
The next step would be find the value of [tex]\z_{\alpha/2}$ , $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$