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kykrilka [37]
3 years ago
14

The vertex of the parabola is at the point (2,-5). which of the equations below could be the one for this parabola?

Mathematics
2 answers:
timofeeve [1]3 years ago
7 0
Hello,
Answer is Dsince y=k(x-2)²-5  and if k=1 ==>D.

Elodia [21]3 years ago
6 0

Answer:

The correct option is D. y=(x-2)^{2}-5

Step-by-step explanation:

If the vertex V of the parabola has the following coordinates :

V=(xV,yV)

One way to write the equation of the parabola is :

y=a(x-xV)^{2}+yV

Where xV and yV are the coordinates from the vertex of the parabola and ''a'' is a real number.

If a>0 ⇒ The parabola is concave upward

If a ⇒ The parabola is concave downward

In this exercise

V=(xV,yV)=(2,-5) ⇒

One way to write this parabola is

y=a(x-2)^{2}-5

The graph of the parabola is concave upward ⇒a>0

The option D. is

y=1(x-2)^{2}-5 Where a=1 and 1>0 ⇒

The correct option is D.

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Step-by-step explanation:

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Determine the number of degrees of freedom for the two-sample t test or CI in each of the following situations. (Round your answ
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Answer:

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Part b ) The degrees of freedom for the given two sample non-pooled t test is 30

Part c ) The degrees of freedom for the given two sample non-pooled t test is 30

Part d ) The degrees of freedom for the given two sample non-pooled t test is 25

Step-by-step explanation:

Degrees of freedom for a non-pooled two sample t-test is given by;

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Now given the information;

a) :- m = 12, n = 15, s₁ = 4.0, s₂ = 6.0

we substitute

Δf =  {[ 4²/12 + 6²/15 ]²} / {[( 4²/12)²/12-1] + [(6²/15)²/15-1]}

Δf  = 30184 / 1241

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b) :- m = 12, n = 21, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/12 + 6²/21 ]²} / {[( 4²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 56320 / 1871

Δf = 30.1015 ≈ 30 (down to the nearest whole number)

c) :- m = 12, n = 21, s₁ = 3.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 3²/12 + 6²/21 ]²} / {[( 3²/12)²/12-1] + [(6²/21)²/21-1]}

Δf = 29095 / 949

Δf = 30.6585 ≈ 30 (down to the nearest whole number)

d) :- m = 10, n = 24, s₁ = 4.0, s₂ = 6.0

we substitute using same formula

Δf = {[ s₁²/m + s₂²/n ]²} / {[( s₁²/m)²/m-1] + [(s₂²/n)²/n-1]}

Δf = {[ 4²/10 + 6²/24 ]²} / {[( 4²/10)²/10-1] + [(6²/24)²/24-1]}

Δf = 1044 / 41  

Δf = 25.4634 ≈ 25 (down to the nearest whole number).

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Answer:

Hope that helps

Step-by-step explanation:

Brainliest would be much appreciated

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