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Ket [755]
3 years ago
12

b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone​ e

lse? A. ​Yes, because 0.01less than<​P(never)less than<0.10. B. ​Yes, because ​P(never)less than<0.05. C. ​No, because there were 139139 people in the survey who said they never wear their seat belt. D. ​No, because the probability of an unusual event is 0.
Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

Answer:

Step-by-step explanation:

Hello!

<em>Full text</em>

<em>In a national survey college students were​ asked, "How often do you  wear a seat belt when riding in a car driven by someone​ else?" The response frequencies appear in the table to the right.​ (a) Construct a probability model for​ seat-belt use by a passenger.​ (b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone​ else? </em>

<em>Response , Frequency   </em>

<em>Never  102 </em>

<em>Rarely  319 </em>

<em>Sometimes  524 </em>

<em>Most of the time  1067 </em>

<em>Always  2727 </em>

n= 102+319+524+1067+2727= 4739

<em> ​(a) Complete the table below. </em>

<em>Response </em>

<em>Probability  </em>To calculate the probability for each response you have to divide the frequency of each category by the total of people surveyed:

<em>Never </em> P(N)= 102/4739= 0.0215

<em>​(Round to the nearest thousandth as​ needed.) </em>

<em>Rarely </em> P(R)= 319/4739= 0.0673

<em>​(Round to the nearest thousandth as​ needed.) </em>

<em>Sometimes </em> P(S)= 524/4739= 0.1106

<em>​(Round to the nearest thousandth as​ needed.) </em>

<em>Most of the time  </em>P(M)= 1067/4739= 0.2252

<em>​(Round to the nearest thousandth as​ needed.) </em>

<em>Always  </em>P(A)= 2727/4739= 0.5754

<em>​(Round to the nearest thousandth as​ needed.) </em>

<em />

<em>​(b) Would you consider it unusual to find a college student who never wears a seat belt when riding in a car driven by someone​ else? </em>

<em>A. </em>

<em>​No, because there were 102  people in the survey who said they never wear their seat belt. </em> Incorrect, an event is considered unusual if its probability (relative frequency) is low, you cannot know if it is usual or unusual just by looking at the absolute frequency of it.

<em>B. </em>

<em>​Yes, because ​P(never) < 0.05. </em> Correct

<em> C. </em>

<em>​No, because the probability of an unusual event is 0.    </em>Incorrect, the probability of unusual events is low, impossible events are the ones with probability zero

<em>D. </em>

<em>​Yes, because 0.01 < ​P(never) <  0.10. </em>Incorrect, by the definition an event is considered unusual when its probability is equal or less than 5%.

I hope this helps!

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The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

  1. Neither
  2. ║
  3. Neither
  4. ⊥
  5. ║
  6. Neither
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Reason:

The slope and intercept form is the form y = m·x + c

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1. The given equations are in the slope and intercept form

\ y = 3 \cdot x + 1

The slope, m₁ = 3

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2. y = 5·x - 3

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The second equation can be rewritten in the slope and intercept form as follows;

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3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

y = 2 -\dfrac{1}{2}  \cdot x

Slope, m₁ = -\dfrac{1}{2}

y = \dfrac{1}{2}  \cdot x - 2

Slope, m₂ = \dfrac{1}{2}

The slopes

Therefore, m₁ ≠ m₂

m_1 \neq -\dfrac{1}{m_2}

The lines are <u>Neither</u> parallel nor perpendicular

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4. The given equations are;

2·y - x = 2

y = \dfrac{1}{2} \cdot   x +1

m₁ = \dfrac{1}{2}

y = -2·x + 4

m₂ = -2

Therefore;

m_1 \neq -\dfrac{1}{m_2}

Therefore, the lines are <u>perpendicular</u>

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5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, y = \dfrac{3}{4} \cdot x + 3

Second equation, y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}

Therefore, m₁ = m₂, the lines are <u>parallel</u>

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6. The given equations are;

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Therefore, the two equations are <u>neither</u> parallel nor perpendicular

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brainly.com/question/16732089

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