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Sedaia [141]
3 years ago
9

NEED HELP ASAP!! WILL MARK BRAINLIEST IF YOU ANSWER RIGHT!!

Mathematics
1 answer:
kakasveta [241]3 years ago
6 0

Answer:

B

Step-by-step explanation: Trust

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Please help. Solve the triangle. Round ans to the nearest tenth.
White raven [17]

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Answer:

  • C = 21°
  • a = 13.3
  • c = 5.4

Step-by-step explanation:

The third angle can be found from the sum of angles in a triangle.

  A + B + C = 180°

  C = 180° -62° -97°

  C = 21°

__

The remaining side lengths can be found using the Law of Sines.

  a/sin(A) = b/sin(B)

  a = sin(62°)(15/sin(97°)) ≈ 13.34

Similarly, ...

  c/sin(C) = b/sin(B)

  c = sin(21°)(15/sin(97°)) ≈ 5.42

The remaining side lengths are approximately ...

  a ≈ 13.3

  c ≈ 5.4

4 0
3 years ago
What is 1+3+1+1+1+1+1+1+1+1+1+1+1+1+1+1
Gemiola [76]

Step-by-step explanation:

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MARK ME BRAINLIEST THANKS MY ANSWER PLEASE

4 0
3 years ago
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Solve the one-variable equation using the distributive property and properties of equality.
Tanzania [10]

Answer:

a=6

Step-by-step explanation:

-6(2+(6)=-48

-6(2+6)=-48

(-6X2)+(-6+6)=-48

(-12)+(-36)=-48

-48=-48

They are equal. Tell me if I made a mistake. Thanks!

6 0
3 years ago
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2 + 2 = ?<br> This is for easy points!
Setler79 [48]
4 is the answert :))
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3 years ago
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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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