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Luda [366]
2 years ago
10

PLEASE NEED HELP

Mathematics
2 answers:
Firdavs [7]2 years ago
5 0

The last statement "Graph B is a valid density curve because the curve is above the horizontal axis, and the area under the curve is 1." is the correct statement.

A few fundamental principles apply to density curves:

  • A density curve's area beneath it represents probability.
  • A density curve's area under it equals one.
  • Base x height in a uniform density curve equals one.
  • The likelihood that x = a will never occur.
  • The likelihood that x < a is the same as that of x ≤ a.

Following the rules above, we can see from graph B that the curve is above the horizontal axis. That shows that the probabilities are positive which is a necessity. Moreover, the area under the curve must also be 1. That is also satisfied.

Hence, the last statement "Graph B is a valid density curve because the curve is above the horizontal axis, and the area under the curve is 1." is the correct statement.

Learn more about density curves here-

brainly.com/question/18345488

#SPJ10

denis-greek [22]2 years ago
5 0

Answer: Graph B is a valid density curve because the curve is above the horizontal axis, and the area under the curve is 1.

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5 0
3 years ago
Solve the system of linear equations below.
Reika [66]

Answer:

x = 1 , y = 3 thus: A is your Anser

Step-by-step explanation:

Solve the following system:

{2 x + y = 5 | (equation 1)

x + y = 4 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y = 5 | (equation 1)

0 x+y/2 = 3/2 | (equation 2)

Multiply equation 2 by 2:

{2 x + y = 5 | (equation 1)

0 x+y = 3 | (equation 2)

Subtract equation 2 from equation 1:

{2 x+0 y = 2 | (equation 1)

0 x+y = 3 | (equation 2)

Divide equation 1 by 2:

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8 0
3 years ago
Read 2 more answers
6+4*(5-7)^2 need answer with worked out problem
Dennis_Churaev [7]
6+4• (5-7)^2
(5-7)^2
-2^2
4
6+4+4
=14
6 0
3 years ago
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