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Setler79 [48]
4 years ago
5

I WILL GIVE EXTRA POINTS AND BRAINLIEST HELPPP PLEASE!!!! SHOW ALL WORK!!!Solve All 4!!

Mathematics
1 answer:
MA_775_DIABLO [31]4 years ago
7 0

Answer:

#27 is 2x+5°=5

x+59°=59

#28 is x+64°=64

3x+26°= -26

#29 is 2x+17°=17

x+77°=77

#30 is x+97°=97

2x+32°=32

Step-by-step explanation:

remember what i told you +=positive -=negative

hope this helped you out!

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

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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

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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
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