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Darina [25.2K]
3 years ago
9

PLEASE HELP ME!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
liberstina [14]3 years ago
8 0
I think the answer is ;B.316 and 38
aleksley [76]3 years ago
3 0

Answer:

B.  \frac{3}{16} and \frac{3}{8}

Step-by-step explanation:

From the given figure, we see that,

Every partition is of length \frac{1}{16}.

So, we get that,

\frac{1}{16} contains 4 data points.

\frac{3}{16} contains 1 data point.

\frac{6}{16} contains 1 data point.

\frac{8}{16}=\frac{1}{2} contains 2 data points.

\frac{10}{16} contains 2 data points.

\frac{12}{16} contains 1 data point.

Since, the largest gap between two data points is \frac{2}{16} which occurs between the points \frac{3}{16} and \frac{6}{16}=\frac{3}{8}

<h3>Hence, B is the correct option.</h3>
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Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is:
professor190 [17]

Answer:

Remember, if B is a set, R is a relation in B and a is related with b (aRb or (a,b))

1. R is reflexive if for each element a∈B, aRa.

2. R is symmetric if satisfies that if aRb then bRa.

3. R is transitive if satisfies that if aRb and bRc then aRc.

Then, our set B is \{1,2,3,4\}.

a) We need to find a relation R reflexive and transitive that contain the relation R1=\{(1, 2), (1, 4), (3, 3), (4, 1)\}

Then, we need:

1. That 1R1, 2R2, 3R3, 4R4 to the relation be reflexive and,

2. Observe that

  • 1R4 and 4R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 4R1 and 1R2, then 4 must be related with 2.

Therefore \{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(4,1),(4,2)\} is the smallest relation containing the relation R1.

b) We need a new relation symmetric and transitive, then

  • since 1R2, then 2 must be related with 1.
  • since 1R4, 4 must be related with 1.

and the analysis for be transitive is the same that we did in a).

Observe that

  • 1R2 and 2R1, then 1 must be related with itself.
  • 4R1 and 1R4, then 4 must be related with itself.
  • 2R1 and 1R4, then 2 must be related with 4.
  • 4R1 and 1R2, then 4 must be related with 2.
  • 2R4 and 4R2, then 2 must be related with itself

Therefore, the smallest relation containing R1 that is symmetric and transitive is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

c) We need a new relation reflexive, symmetric and transitive containing R1.

For be reflexive

  • 1 must be related with 1,
  • 2 must be related with 2,
  • 3 must be related with 3,
  • 4 must be related with 4

For be symmetric

  • since 1R2, 2 must be related with 1,
  • since 1R4, 4 must be related with 1.

For be transitive

  • Since 4R1 and 1R2, 4 must be related with 2,
  • since 2R1 and 1R4, 2 must be related with 4.

Then, the smallest relation reflexive, symmetric and transitive containing R1 is

\{(1,1),(2,2),(3,3),(4,4),(1,2),(1,4),(2,1),(2,4),(3,3),(4,1),(4,2),(4,4)\}

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3 years ago
Bags of nails weigh 200 grams each.
Tems11 [23]

Answer:ewlkdgvs8u9ejrmxc

Step-by-step explanation:lu7iy 6t5r47oe;8jo/ilni.ub8yu

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3 years ago
X squared plus 8 squared equals 9<br> What is X?
BabaBlast [244]

Answer:

x = i\sqrt{55} , -i\sqrt{55}

Step-by-step explanation:

x^2 + 8^2 = 9

Simplify the 8^2

x^2 + 64 = 9

Subtract 64 from both sides

x^2 = -55

Take the square root from both sides

x = +/- \sqrt{-55}

Simplify

x = i\sqrt{55} , -i\sqrt{55}

Hope this helps!

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