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2 years ago

Tyrell is going to be use ASA to prove that PQR=SQR

Mathematics

2 answers:

motikmotik2 years ago

8 0

Presumably our goal is to show PRQ=SRQ, PQR=SQR and QR=QR for ASA.

A. QR=QR is the reflexive property; things are congruent to themselves, TRUE

B. We need to prove that, but it doesn't have anything to do with the symmetric property.

C. PQ=SQ isn't something we'd need to show for our ASA proof

D. Again, not the symmetric property.

Answer: A

N76 [4]2 years ago

7 0

Answer: A. Prove that QR≅QR by the reflexive property.

Step-by-step explanation:

ASA postulate says that if two angles and the included side of a triangle are congruent to two angles and the included side of other triangle then the triangles are said to be congruent.

In the given triangles ΔPQR and ΔSQR

∠PQR ≅∠SQR {given}

∠QRP ≅ ∠QRS {given}

In the figure , the included side of ∠PQR and ∠QRP= QR

The included side of ∠SQR and ∠QRS =OR

So we need to probe QR≅QR by Reflexive property, to prove triangles ΔPQR and ΔSQR are congruent.

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2 years ago

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so percentage of students who chose art = 52%

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2 years ago

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A regular hexagon rotates counterclockwise around its center. It turns through angles greater than 0° and less or equal to 360°.

Sonja [21]

there are 6 sides to a hexagon

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2 years ago

Which expression is equivalent to the expression below? StartFraction m + 3 Over m squared minus 16 EndFraction divided by Start

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Option B: $\frac{1}{(m-4)(m-3)}$ is the correct answer.

Explanation:

The given expression is $\frac{(\frac{m+3}{m^2-16}) }{(\frac{m^2-9}{m+4} )}$

Simplifying the expression, we have,

$\frac{m+3}{m^{2}-16}\times\frac{m+4}{m^2-9}$

Factor the equations, $m^{2}-16\right$ and $m^{2}-9$ ,we get,

$m^{2}-16\right=m^{2}-4^{2}=(m+4)(m-4)$

$m^{2}-9=m^{2}-3^2=(m+3)(m-3)$

Substituting these factored expressions in the above expression, we have,

$\frac{m+3}{(m+4)(m-4)}\times\frac{m+4}{(m+3)(m-3)}$

Cancelling the common terms $m+3$ and $m+4$ , we get,

$\frac{1}{(m-4)(m-3)}$

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Hence, Option B is the correct answer.

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2 years ago

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Answer:

At the 5% level, BAOI can infer that the average time to complete does not exceeds 60 hours.

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 60 \ hr$

The sample size is $n = 16$

The sample mean is $\= x = 68 \ hr$

The standard deviation is $\sigma = 20 \ hr$

The null hypothesis is $H_o : \mu = 60$

The alternative $H_a : \mu > 60$

Here we would assume the level of significance of this test to be

$\alpha = 5\% = 0.05$

Next we will obtain the critical value of the level of significance from the normal distribution table, the value is $Z_{0.05} = 1.645$

Generally the test statistics is mathematically represented as

$t = \frac{ \= x - \mu}{ \frac{ \sigma }{\sqrt{n} } }$

substituting values

$t = \frac{ 68 - 60 }{ \frac{ 20 }{\sqrt{16} } }$

$t = 1.6$

Looking at the value of t and $Z_{\alpha }$ we see that $t< Z_{\alpha }$ hence we fail to reject the null hypothesis

This means that there no sufficient evidence to conclude that it takes more than 60 hours to complete the course

At the 5% level, BAOI can infer that the average time to complete does not exceeds 60 hours.

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