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Ilia_Sergeevich [38]
3 years ago
7

Y=5x+2 4x−y=0 ​ Is (2,12) a solution of the system?

Mathematics
2 answers:
STALIN [3.7K]3 years ago
8 0

Answer:

no the solution would be (-2, -8)

Step-by-step explanation:

i graphed both of the equations and found the solution <33

Anvisha [2.4K]3 years ago
8 0

Answer:

(-2,-8) is correct

Step-by-step explanation:

4x = y

5x +2 = y

4x = 5x +2

-2 = x

y = 5×-2 + 2

=-8

( x , y )

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Answer:

a) The 95% confidence interval would be given by -1.887 \leq \mu_1 -\mu_2 \leq 2.687  

b) No contradict the result obtained since the confidence interval contains the value 0, so we don't have enough evidence to conclude that we have a significant effect in the colling rate and the hardness.

Step-by-step explanation:

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".  

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\bar X_1 =91.1 represent the sample mean 1

\bar X_2 =90.7 represent the sample mean 2

n1=50 represent the sample 1 size  

n2=40 represent the sample 2 size  

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The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} (1)  

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In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:  

df=n_1 +n_2 -1=50+40-2=88  

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,88)".And we see that t_{\alpha/2}=1.987  

The standard error is given by the following formula:

SE=\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}

And replacing we have:

SE=\sqrt{\frac{6.55^2}{50}+\frac{4.32^2}{40}}=1.325

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0.4+1.987\sqrt{\frac{6.55^2}{50}+\frac{4.32^2}{40}}=2.687  

So on this case the 95% confidence interval would be given by -1.887 \leq \mu_1 -\mu_2 \leq 2.687  

Part b

No contradict the result obtained since the confidence interval contains the value 0, so we don't have enough evidence to conclude that we have a significant effect in the colling rate and the hardness.

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