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djyliett
2 months ago
6

Two objects are moving along separate linear paths where each path is described by position, d, and time, t. The variable d is m

easured in meters, and the variable t is measured in seconds. The equation describing the graph of the position of the first object with respect to time is d = 2.5t + 2.2. The graph of the position of the second object is a parallel line passing through (t = 0, d = 1). What is the equation of the second graph?
Mathematics
2 answers:
lawyer [12.5K]2 months ago
8 0
I’m not entirely certain, but here’s my attempt: d=2.5t+2.2 closely resembles y=mx+b with y=d, m=2.5, x=t, and b=2.2. To derive a new equation, we replace d and t, resulting in 1=2.5(0)+b. After simplification, we find b=1, so the new equation becomes d=2.5t+1
.
Inessa [12.5K]2 months ago
3 0

Response:

The equation representing the second graph is:

d=2.5t+1

Step-by-step explanation:

Just as a reminder, the slope of a line that is parallel to another maintains the same slope.

In this case, the slope for the second line is 2.5

( As the first line has a slope of 2.5 )

Additionally, we know the formula for the equation of a line given a point (a,b) and slope m is:

y-b=m(x-a)

Where (a,b)= (0,1) and m=2.5, y=d, and x=t

Thus, the resulting equation for the second line is given by:

d-1=2.5(t-0)\\\\\\d-1=2.5t\\\\d=2.5t+1

.
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Answer:

Answer and Explanation:

We have:

Population mean,

μ

=

3

,

000

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Population standard deviation,

σ

=

696

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Sample size,

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36

1) The standard deviation for the sampling distribution:

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2) By the central limit theorem, the sampling distribution's expected value matches the population mean.

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The expected value of the sampling distribution equals the population mean,

μ

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μ

=

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,

000

The standard deviation of the sampling distribution,

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¯

x

=

116

The sampling distribution of

¯

x

is roughly normal due to a sample size greater than

30

.

3) The likelihood that the average lifespan of the sample falls between

2670.56

and

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P

(

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<

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<

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=

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−

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<

z

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3000

116

)

=

P

(

−

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<

z

<

−

1.64

)

=

P

(

z

<

−

1.64

)

−

P

(

z

<

−

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=

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In Excel: =NORMSDIST(-1.64)-NORMSDIST(-2.84)

4) The probability of the average life in the sample exceeding

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P

(

x

>

3219.24

)

=

P

(

z

>

3219.24

−

3000

116

)

=

P

(

z

>

1.89

)

=

0.0294

In Excel: =NORMSDIST(-1.89)

5) The likelihood that the sample's average life is lower than

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P

(

x

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3180.96

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=

P

(

z

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3000

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)

=

P

(

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