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PIT_PIT [208]
3 years ago
10

pectrum Inc., an automobile manufacturer, is interested in measuring the customer satisfaction of car owners. It decides to coll

ect data from new purchasers of Spectrum cars. In this scenario, the new purchasers of Spectrum cars represent a(n) _____. sampling unit construct estimated sample deviation sampling frame estimated sample variance
Physics
1 answer:
ki77a [65]3 years ago
6 0

Answer:

Sampling unit.

Each customer from whom information is collected is a sample unit which when gathered together in large numbers becomes a data

Explanation:

The accumulation of sampling units is what makes a data, sampling unit is an individual value within a sample's database.

It is also the list of individual sampling units with proper identification and features which are to be measured during analysis. A sampling unit is also called elementary unit

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Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w
Tatiana [17]

Answer:

The average velocity is

266\frac{m}{s},274\frac{m}{s} and 117\frac{m}{s} respectively.

Explanation:

Let's start writing the vertical position equation :

L(t)=2t^{3}+t^{2}-5t+1

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

V_{avg}=\frac{Displacement}{Time} = Δx / Δt = \frac{x2-x1}{t2-t1}

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

t2-t1=8s-5s=3s

For the position variation we use the vertical position equation :

x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m

x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

\frac{798m}{3s}=266\frac{m}{s}

For the second time interval :

t1 = 4 s → t2 = 9 s

x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m

x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

\frac{1370m}{5s}=274\frac{m}{s}

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m

x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

\frac{702m}{6s}=117\frac{m}{s}

5 0
3 years ago
Calculate the displacement of the volleyball in sample problem 2f when teh volleyballs final velocity is 1.1 m/s upward
stich3 [128]
We have: v i (initial velocity) = 6 m/sv = 1.1 m/sa = - 9.8 m/s²v = v i + a · t1.1 m/s = 6 m/s - 9.8 m/s² t9.8 t = 6 - 1.19.8 t = 4.9t = 4.9 : 9.8t = 0.5 sThen the replacement:x = xi + vi · t + a t² / 2( xi = 0 )x = 6 · 0.5 - 9.8 · 0.25 / 2x = 3 - 1.225Answer:
x = 1.775 m

4 0
4 years ago
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