Related questions with answers

Television advertising would ideally be aimed at exactly the audience that observes the ads. A study was conducted to determine the amount of time that individuals spend watching TV during evening prime-time hours. Twenty individuals were observed for a 1-week period, and the average time spent watching TV per evening, Y , was recorded for each. Four other bits of information were also recorded for each individual:

x1=agex _ { 1 } = a g e

,

x2= education level. x _ { 2 } = \text { education level. }

,

x3= disposable income x _ { 3 } = \text { disposable income }

, and

x4=IQx _ { 4 } = \mathrm { I } \mathrm { Q }

. Consider the three models given below:

 Model I: Y=β0+β1x1+β2x2+β3x3+β4x4+ε\text { Model I: } \quad Y = \beta _ { 0 } + \beta _ { 1 } x _ { 1 } + \beta _ { 2 } x _ { 2 } + \beta _ { 3 } x _ { 3 } + \beta _ { 4 } x _ { 4 } + \varepsilon

 Model II: Y=β0+β1x1+β2x2+ε\text { Model II: } Y = \beta _ { 0 } + \beta _ { 1 } x _ { 1 } + \beta _ { 2 } x _ { 2 } + \varepsilon

 Model III: Y=β0+β1x1+β2x2+β3x1x2+ε\text { Model III: } Y = \beta _ { 0 } + \beta _ { 1 } x _ { 1 } + \beta _ { 2 } x _ { 2 } + \beta _ { 3 } x _ { 1 } x _ { 2 } + \varepsilon

Are the following statements true or false? Model II is a reduction of model I.

Question

Escape velocity from the Earth is 11.2 km/s11.2 \mathrm{~km} / \mathrm{s}. What would be the percent decrease in length of a 73.6 m73.6\mathrm{~m} long spacecraft traveling at that speed as seen from Earth?

Solution

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Information

In this task, we know the ship's length:

  • l0=73.6 m\mathrm{l_0=73.6\ m}

We need to calculate the percentage decrease if the ship travels with speed:

  • v=11.2 kms=11.2103 ms\mathrm{v=11.2\ \frac{km}{s}=11.2 \cdot 10^3\ \frac{m}{s}}

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