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melamori03 [73]
3 years ago
5

Thinking about the areas of a Standard Normal Distribution, as z values decrease, do the areas to the left of z increase, or dec

rease?

Mathematics
1 answer:
Vera_Pavlovna [14]3 years ago
8 0
The figure below shows the standard normal distribution or "bell-shaped" curve, plotted against the z-score.
The z-score is defined as
z = (x - μ)/σ
where
x = nrandom variable
μ = mean
σ = standard deviation.

As z-values decrease, areas to the left of z decrease as shown by the shaded area.

Answer:
Areas to the left of z decrease.

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Ad libitum [116K]

Answer:

A and D

Step-by-step explanation:

53 + 37 + 44 with the time standard being minutes

Find total time

___________________________________________

We have units of time and need to add them together, let's do as such.

53 + 37 + 44

134 minutes.

Let's try simplifying it into minutes to find how much we have :

134/60

2 hours and 14 minutes

Therefore A and D are correct.

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2 years ago
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Find a such that the solution of y'' + y' − 2y = 0, y(0) = a, y' (0) = 1 tends to zero as t → +∞.
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Answer:

Step-by-step explanation:

y'' + y' − 2y = 0, y(0) = a, y' (0) = 1

Auxialary equation is

m^2+m-2=0\\m=-2,1

General solution is

y=Ae^{-2x} +Be^x

y(0) = A+B =a

y'(0) = -2Ae^{2x} +Be^x = -2A+B = 1

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3A =a-1

We know that y tends to 0 when x tends to infinity for any finite A

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3 years ago
Find the volume of the given solid. Under the surface z = 6xy and above the triangle with vertices (1, 1), (4, 1), and (1, 2)
Leya [2.2K]

he volume of the solid under a surface  

z

=

f

(

x

,

y

)

and above a region D is given by the formula  

∫

∫

D

f

(

x

,

y

)

d

A

.

Here  

f

(

x

,

y

)

=

6

x

y

. The inequalities that define the region D can be found by making a sketch of the triangle that lies in the  

x

y

−

plane. The bounding equations of the triangle are found using the point-slope formula as  

x

=

1

,

y

=

1

and  

y

=

−

x

3

+

7

3

.

Here is a sketch of the triangle:

Intersecting Region

The inequalities that describe D are given by the sketch as:  

1

≤

x

≤

4

and  

1

≤

y

≤

−

x

3

+

7

3

.

Therefore, volume is

V

=

∫

4

1

∫

−

x

3

+

7

3

1

6

x

y

d

y

d

x

=

∫

4

1

6

x

[

y

2

2

]

−

x

3

+

7

3

1

d

x

=

3

∫

4

1

x

[

y

2

]

−

x

3

+

7

3

1

d

x

=

3

∫

4

1

x

[

49

9

−

14

x

9

+

x

2

9

−

1

]

d

x

=

3

∫

4

1

40

x

9

−

14

x

2

9

+

x

3

9

d

x

=

3

[

40

x

2

18

−

14

x

3

27

+

x

4

36

]

4

1

=

3

[

(

640

18

−

896

27

+

256

36

)

−

(

40

18

−

14

27

+

1

36

)

]

=

23.25

.

Volume is  

23.25

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4 0
3 years ago
Can someone help me please
sergeinik [125]
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