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Zepler [3.9K]
3 years ago
6

Given: ∆ABC, AB = CB BD − median to AC E∈ AB ,F∈ BC AE = CF Prove: △ADE ≅ △CDF ΔBDE ≅ ΔBDF

Mathematics
1 answer:
NemiM [27]3 years ago
7 0

Answer:

1) By SAS theorem, ΔADE≅ΔCDF

2) By SSS theorem, ΔBDE≅ΔBDF

Step-by-step explanation:

Consider isosceles triangle ABC (see diagram).

1. In triangles ADE and CDF:

  • AD≅DC (since BD is median, then it divides side AC in two congruent parts);
  • AE≅CF (given);
  • ∠A≅∠C (triangle ABC is isosceles, then angles adjacent to the base are congruent).

By SAS theorem, ΔADE≅ΔCDF.

2. In triangles BDE and BDF:

  • side BD is common;
  • DE≅DF (ΔADE≅ΔCDF, then congruent triangles have congruent corresponding sides);
  • BE≅FB (triangle ABC is isosceles, AB≅BC, AE≅CF, then BE=AB-AE, FB=BC-CF).

Be SSS theorem, ΔBDE≅ΔBDF.

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Suppose 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance
lana66690 [7]

Answer:

Null Hypothesis, H_0 : p = 0.20  

Alternate Hypothesis, H_a : p > 0.20  

Step-by-step explanation:

We are given that 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea.

We have to use a 0.05 significance level to test the claim that more than 20​% of users develop nausea.

<em>Let p = population proportion of users who develop nausea</em>

So, <u>Null Hypothesis,</u> H_0 : p = 0.20  

<u>Alternate Hypothesis</u>, H_a : p > 0.20  

Here, <u><em>null hypothesis</em></u> states that 20​% of users develop nausea.

And <u><em>alternate hypothesis</em></u> states that more than 20​% of users develop nausea.

The test statistics that would be used here is <u>One-sample z proportion</u> test statistics.

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mariarad [96]

The Lena has no common ratio for the sequence 1.0 1.5 2.5 2.0

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