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seropon [69]
3 years ago
7

Based on △GHI below, which of the following inequalities is true?

Mathematics
1 answer:
Rom4ik [11]3 years ago
4 0

Answer:

C. \overline{GI} < \overline{GH} < \overline{HI}

Step-by-step explanation:

In a triangle, the longest side is opposite to the largest measure of angle. The medium side is opposite to the medium angle, while the shortest side is opposite to the smallest angle.

In ∆GHI,

✍️m<G = 95°, it is opposite to \overline{HI}

<G is the largest angle, therefore, \overline{HI} is the longest side.

✍️m<I = 180 - (95 + 38) = 47°.

<I is opposite to \overline{GH}

<I is the medium angle, therefore, \overline{GH} is the medium side.

✍️m<H = 38°, it is opposite to \overline{GI}

<h is the smallest angle, therefore, \overline{GI} is the shortest side.

✅This means:

\overline{GI} < \overline{GH} < \overline{HI}

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3 years ago
Help? 2+2-8+6-2+9+5-6+6-3+8-1+7-0+8-1+10-2+11+
SOVA2 [1]

Answer:

51

Step-by-step explanation:

2+2 = 4

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6 0
3 years ago
student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te
alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = \frac{1}{4}.

Thus, the probability that the student does not receive version A = 1-\frac{1}{4} = \frac{3}{4}.

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

= (\frac{3}{4^4})[1+\frac{1}{16}]

= (\frac{3}{256})[\frac{17}{16}]

= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

4 0
3 years ago
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The answer is -3, hope this helps!

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