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drek231 [11]
1 year ago
7

Kurtosis of a normal data distribution is a ___________________ Group of answer choices Measure of data centrality Measure of da

ta quality Measure of data shape Measure of data dispersion
Mathematics
1 answer:
Irina18 [472]1 year ago
5 0

Answer:

Option A, Measure of data centrality

Step-by-step explanation:

Normal distribution curve determines the skewness of the data as to what extent is is disoriented from the normal distribution curve. Also, it is the probability distribution curve that determines whether the data set is two legged, one legged or it lies under which probability curve.

Hence option A is correct

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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)\}

5 0
1 year ago
Which congruence theorem can be used to prove △BDA ≅ △DBC? Triangles B D A and D B C share side D B. Angles C B D and A D B are
Tomtit [17]

Answer:

A. Hypotenuse-leg (HL) congruence.

HL, when you have 2 right triangles and their hypotenuses are congruent you are able to say HL

Step-by-step explanation:

We know that the hypotenuse-leg theorem states that if the hypotenuse and one leg of a right triangle are congruent to hypotenuse and corresponding leg of another right triangle, then the triangles are congruent.  

hypotenuse(AB) of △BDA equals to hypotenuse (CD) of △DBC.  

BDA and DBC share a common side DB.

Using Pythagorean theorem we will get,

CD^{2}=DB^{2}+BC^{2}...(1)  \\\\AB^{2}=DB^{2}+AD^{2}...(2)

We have been given that CD=AB, Upon using this information we will get,

DB^{2}+BC^{2}=DB^{2}+AD^{2}

Upon subtracting DB^{2} from both sides of our equation we will get,

BC^{2}=AD^{2}\\\\BC=AD

<h3>Therefore, by HL congruence △BDA ≅ △DBC.</h3>
6 0
1 year ago
Read 2 more answers
Solve for x. −12 = − 3/4 x A) 6 B) 8 C) 12 D) 16
shepuryov [24]

Answer:

D) 16

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
In △ABC, m∠ABC=40°,
Alchen [17]

Let m∠CLN = x. Then m∠ALM = 3x, and m∠A = 90°-x, m∠C = 90°-3x.

The sum of angles of ∆ABC is 180°, so we have

... 180° = 40° + m∠A + m∠C

Using the above expressions for m∠A and m∠C, we can write ...

... 180° = 40° + (90° -x) + (90° -3x)

... 4x = 40° . . . . . . . . . add 4x-180°

... x = 10°

From which we conclude ...

... m∠C = 90°-3x = 90° - 3·10° = 60°

The ratio of CN to CL is

... CN/CL = cos(∠C) = cos(60°)

... CN/CL = 1/2

so ...

... CN = (1/2)CL

5 0
2 years ago
5/8 of the staff are male 5/12 of the staff works part time at the aquarium what is the fraction of the staff being female
Goryan [66]
I believe the answer is 3/8. The whole portion of employees, which is translated as 8/8 is deducted by 5/8, which is the population of male employees.
7 0
2 years ago
Read 2 more answers
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