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Sergeu [11.5K]
3 years ago
9

Find the volume and total surface area of a regular tetrahedron whose measure of one edge is 10cm

Mathematics
2 answers:
mixas84 [53]3 years ago
7 0
The total surface area of a regular tetrahedron is calculated through the equation,
                            SA = (sqrt 3) x a²
Substituting the known value,
                            SA = (sqrt 3) x (10 cm)² = 173.20 cm²
Its volume is obtained by,
                             V = ((sqrt 2)/ 12) x a³
                         V = ((sqrt 2) / 12) x (10 cm)³ = 117.85 cm³
Thus, its surface area and volume are 173.20 cm² and 117.85 cm³, respectively.
Diano4ka-milaya [45]3 years ago
3 0
For a tetrahedron, the total surface area is expressed as:
SA = √3 x a²
Substituting the given value,
SA = √3 x (10 cm)² = 173.20 cm²

The volume is expressed as,
V = (√2/ 12) x a³

Substituting the given values,
V = (<span>√2</span> / 12) x (10 cm)³ = 117.85 cm³
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Marina CMI [18]

Answer:

The median of A is the same as the median of B.

The interquartile range of B is greater than the interquartile range of A.

Step-by-step explanation:

Given that:

A = number of runs allowed in first 9 games

A = {1, 4, 2, 2, 3, 1, 1, 2, 1}

Rearranging A : 1, 1, 1, 1, 2, 2, 2, 3, 4

Median A = 1/2(n + 1) th term

Median A = 1/2(10) = 5th term = 2

Q1 of A = 1/4(10) = 2.5th term = (1 + 1)/ 2 = 1

Q3 of A = 3/4(10) = 7.5th term = (2+3)/2 = 2.5

Interquartile range = Q3 - Q1 = 2.5 - 1 = 1.5

Number of runs allowed in 10th game = 9

B = {1, 4, 2, 2, 3, 1, 1, 2, 1, 9}

Rearranging B = 1, 1, 1, 1, 2, 2, 2, 3, 4, 9

Median A = 1/2(n + 1) th term

Median A = 1/2(11) = 5.5th term = (2+2)/2 = 2

Q1 of A = 1/4(11) = 2.75th tetm = (1 + 1)/ 2 = 1

Q3 of A = 3/4(11) = 8.25th term = (3+4)/2 = 3.5

Interquartile range = Q3 - Q1 = 3.5 - 1 = 2.5

Median A = 2 ; median B = 2

IQR B = 2.5 ; IQR A = 1.5 ; IQR B > IQR A

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2 years ago
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