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stira [4]
3 years ago
10

The congruence theorem that can be used to prove ALON

Mathematics
1 answer:
Arisa [49]3 years ago
3 0

Answer:

SSS is the congruence theorem that can be used to prove Δ LON  is congruent to Δ LMN ⇒ 1st answer

Step-by-step explanation:

Let us revise the cases of congruence

  • SSS ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ
  • SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and including angle in the 2nd Δ
  • ASA ⇒ 2 angles and the side whose joining them in the 1st Δ ≅ 2 angles and the side whose joining them in the 2nd Δ
  • AAS ⇒ 2 angles and one side in the 1st Δ ≅ 2 angles and one side in the 2nd Δ
  • HL ⇒ hypotenuse leg of the 1st right Δ ≅ hypotenuse leg of the 2nd right Δ  

In triangles LON and LMN

∵ LO ≅ LM ⇒ given

∵ NO ≅ NM ⇒ given

∵ LN is a common side in the two triangles

- That means the 3 sides of Δ LON are congruent to the 3 sides

   of Δ LMN

∴ Δ LON ≅ LMN ⇒ by using SSS theorem of congruence

SSS is the congruence theorem that can be used to prove Δ LON  is congruent to Δ LMN

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Step-by-step explanation:

Data given and notation  

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We need to conduct a hypothesis in order to check if the mean is equal to 12, the system of hypothesis would be:  

Null hypothesis:\mu \leq 12  

Alternative hypothesis:\mu > 12  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

z=\frac{12.95-12}{\frac{3}{\sqrt{36}}}=1.9  

P-value  

Since is a right tailed test the p value would be:  

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Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 12 at 5% of signficance.  

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