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2 years ago

CHANGE TO STANDARD FORM y = -2/5x + 1/10 y = -3x

Mathematics

1 answer:

Zarrin [17]2 years ago

3 0

Y=-2/5x+1/10=4x+10y=10
y=-3x=3x+y=0
hope these answers help

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Answer:

3rd option

Step-by-step explanation:

$\frac{3x^2-3}{x^2-5x+4}$ ( factorise numerator and denominator )

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Read 2 more answers

Calculating the degrees of freedom, the sample variance, and the estimated standard error for evaluations using the t statistic

Yuliya22 [10]

Answer:

a) For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

$df = n-1= 10-1=9$

d.9

b) $s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15$

a.15

c) For this case we have the sample size n = 25 and the sample variance is $s^2 =400$ , the standard error can founded with this formula:

$SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80$

Step-by-step explanation:

Part a

For the first part we have a sample of n =10 and we want to find the degrees of freedom, and we can use the following formula:

$df = n-1= 10-1=9$

d.9

Part b

From a sample we know that n=41 and SS= 600, where SS represent the sum of quares given by:

$SS = \sum_{i=1}^n (X_i -\bar X)^2$

And the sample variance for this case can be calculated from this formula:

$s^2 = \frac{SS}{n-1}= \frac{600}{41-1}= 15$

a.15

Part c

For this case we have the sample size n = 25 and the sample variance is $s^2 =400$ , the standard error can founded with this formula:

$SE = \frac{s^2}{\sqrt{n}}= \frac{400}{\sqrt{25}}= 80$

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2 years ago