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sp2606 [1]
3 years ago
11

If the slope of line p is 1/5 and the slope of line q is -5 then what is true about lines p and q? (assume lines p and q are cop

lanar)
a. they are parallel
b. they are perpendicular
c. they are neither parallel nor perpendicular
d. they are skew lines
Mathematics
2 answers:
Snowcat [4.5K]3 years ago
8 0

Answer:

B. They are perpendicular

Step-by-step explanation:

They fit the criteria for what you need of a set of perpendicular lines.

vladimir2022 [97]3 years ago
6 0

For this case, we have to define perpendicular lines:

m1 * m2 = -1

We have then:

m1 = \frac {1} {5}\\m2 = -5

Then, according to the definition we have:

(\frac {1} {5}) (- 5) = \frac {-5} {5} = -1

We observe that the definition is fulfilled, therefore, the lines are perpendicular.

Answer:

b. they are perpendicular

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c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

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Gas mileage A: Mean 36, standard deviation 6, sample of 50:

So

\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485

Gas mileage B: Mean 42, standard deviation 8, sample of 50:

So

\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314

Distribution of the difference:

Mean:

\mu = \mu_A - \mu_B = 36 - 42 = -6

Standard error:

s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142

A. Find the point estimate.

This is the difference of means, that is, -6 mpg.

B. Find the margin of error

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

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M = zs = 1.96*1.4142 = 2.77

The margin of error is of 2.77 mpg

C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)

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The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg

The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

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