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hammer [34]
2 years ago
14

Help please!! 20 points and brainliest!!

Mathematics
1 answer:
PSYCHO15rus [73]2 years ago
4 0
Alright, so we have
<span>-9x+6y=-30
</span><span>-12x+9y=-30

Although we might really want to say -9x+6y=-12x+9y, we can't do it that way. We have to get rid of a variable and go from there. Solving for y, we get
</span>-9x+6y=-30
Divide both sides by 3
-3x+2y=-10
Add 3x to both sides
2y=-10+3x
Divide both sides by 2
y=(-10+3x)/2

-12x+9y=-30
Divide both sides by 3
-4x+3y=-10
Add 4x to both sides
-10+4x=3y
Divide both sides by 3
(-10+4x)/3=y


Since we have 
(-10+4x)/3=y and y=(-10+3x)/2 , we have
(-10+4x)/3=(-10+3x)/2
Since 3*2=6, we can multiply both sides by 6
-20+8x=-30+9x
Add 30 to both sides
10+8x=9x
Subtract 8x from both sides
x=10
Plug that in to any equation,and you should have your answer! Good luck, and ask if you have any questions!

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3 years ago
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Setler [38]
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Is the following definition of perpendicular reversible? Two lines that intersect at right angles are perpendicular.
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7 0
3 years ago
An independent-measures research study was used to compare two treatment conditions with n= 12 participants in each treatment. T
Maurinko [17]

Answer:

(a) The data indicate a significant difference between the two treatments.

(b) The data do not indicate a significant difference between the two treatments.

(c) The data indicate a significant difference between the two treatments.

Step-by-step explanation:

Null hypothesis: There is no difference between the two treatments.

Alternate hypothesis: There is a significant difference between the two treatments.

Data given:

M1 = 55

M2 = 52

s1^2 = 8

s2^2 = 4

n1 = 12

n2 = 12

Pooled variance = [(n1-1)s1^2 + (n2-1)s2^2] ÷ (n1+n2-2) = [(12-1)8 + (12-1)4] ÷ (12+12-2) = 132 ÷ 22 = 6

Test statistic (t) = (M1 - M2) ÷ sqrt [pooled variance (1/n1 + 1/n2)] = (55 - 52) ÷ sqrt[6(1/6 + 1/6)] = 3 ÷ 1.414 = 2.122

Degree of freedom = n1+n2-2 = 12+12-2 = 22

(a) For a two-tailed test with a 0.05 (5%) significance level and 23 degrees of freedom, the critical values are -2.069 and 2.069.

Conclusion:

Reject the null hypothesis because the test statistic 2.122 falls outside the region bounded by the critical values.

(b) For a two-tailed test with a 0.01 (1%) significance level and 23 degrees of freedom, the critical values are -2.807 and 2.807.

Conclusion:

Fail to reject the null hypothesis because the test statistic 2.122 falls within the region bounded by the critical values.

(c) For a one-tailed test with 0.05 (5%) significance level and 23 degrees of freedom, the critical value is 1.714.

Conclusion:

Reject the null hypothesis because the test statistic 2.122 is greater than the critical value 1.714.

6 0
3 years ago
2 096 to the nearest 1000
k0ka [10]
The answer to this is 2000...
hope this helps and HAGD!
6 0
2 years ago
Read 2 more answers
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