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Katyanochek1 [597]
3 years ago
8

What is the distance between the points (21, 16) and (9, 11)?

Mathematics
1 answer:
Colt1911 [192]3 years ago
3 0
The distance would be 12,5
but if its the 21,16 going to 9,11 it would most likely end up being -12,-5
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C. SA = 148.8 ft

Step-by-step explanation:

3.1×8×6= 148.8

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Please help this is due tomorrow help
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Step-by-step explanation:

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The sum of the lengths of any two sides of a triangle must be greater than the third side. if a triangle has one side that is 8
BartSMP [9]

1st Side: 8

2nd Side: 2x - 1

3rd Side: x

(8) + (2x - 1) > x            (2x - 1) + x > 8                        (x) + (8) > 2x - 1

       2x + 7 > x                     3x - 1 > 8                                8  > x - 1

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             x > -3.5 (disregard)       x > 3                                 x < 9

3rd Side: x  ⇒ (3 < x < 9)

2nd Side: 2x - 1 ⇒ 2(3) - 1 < x < 2(9) - 1   ⇒   5 < x < 17

Answer: the 3rd side must be between 3 and 9, the 2nd side must be between 5 and 17


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3 years ago
A rectangle with a length of x+5 has a perimeter of 4x+ 14. What is the width of the rectangle?
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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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