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Harlamova29_29 [7]
3 years ago
8

*Extra points* Determine if the lines are perpendicular, parallel, or neither. ​

Mathematics
2 answers:
My name is Ann [436]3 years ago
5 0

Answer:

Step-by-step explanation:

okay I learned how to do this and here are some tips and tricks

when your trying to find if its parallel,perpendicular,or neither

it is best to find it in slope intercept form. so that way you can compare the two slopes

parallel=same slopes

perpendicular= if you change the sign and reciprocal

neither=it cant be flipped

the first equation is 3x-5y=10

goal:we have to convert this into slope intercept

1)find our x and y intercepts

10/3,0 and 0,2

2)write it in point slope form

y-2=3/5(x-10/3)

y-2=3/5x-2

y=3/5x

3/5 is our slope for the first equation

the second equation:10x+6y=-36

1)find x and y intercepts

0,-6

-18/5,0

now if you want to take a short cut here is what you do

-a/b

so.....

-10/6

=-5/3 is your slope for the 2nd equation

so 3/5 and -5/3

not parallel because it doesn't have the same slope different slopes!

its perpendicular because

if you flip the reciprocal and change the sign it would be 3/5

Katyanochek1 [597]3 years ago
3 0
The lines are perpendicular.



If you simplify the first equation you would get y = -3/5x -2.


If you simplify the second equation you would get y = 5/3x -6 which is the negative reciprocal of the first equation. This means that the lines would be perpendicular.
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Step-by-step explanation:

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Step-by-step explanation:

Hello!

You have two variables of interest

X₁: failure stress of a NiCrAlZr coating after nine 1-hr cycles.

X₂: failure stress of a NiCrAlZr coating after six 1-hr cycles.

a)

To be able to estimate the difference between the means using a confidence interval, you need that both variables have a normal distribution and to determine whether or not the population variances are equal.

If the population variances are equal, σ₁²=σ₂², you can use a pooled variance t-test

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b) and c)

You need to test if both population means are the same, the hypotheses are:

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t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

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The distribution of this test is a t with 13 degrees of freedom and the test is two-tailed, so to calculate the p-value you have to do the following:

P(t₁₃≤-4.23)+P(t₁₃≥4.23)= P(t₁₃≤-4.23)+[1-P(t₁₃<4.23)]=  0.000492 + (1-0.999508)= 2*0.000492= 0.000984≅ 0.001

The p-value: 0.001 is less than α: 0.05, the decision is to reject the null hypothesis.

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