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emmainna [20.7K]
3 years ago
10

Formula to calculate variance​

Mathematics
1 answer:
finlep [7]3 years ago
8 0

Answer:

sample variance

the value of the one observation

the mean value of all observations

the number of observations

Step-by-step explanation:

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-9x + 12x + 10 simplifies to
Free_Kalibri [48]

Answer:

\boxed{\sf 3x + 10}

Step-by-step explanation:

Simplify the following:

⇒-9 x + 12 x + 10

Grouping like terms:

⇒(12 x - 9 x) + 10

12 x - 9 x = 3 x:

⇒3 x + 10

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3 years ago
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How do I solve 1.5 to the -2 power or 1.5^-2
zzz [600]

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Step-by-step explanation: hope this help

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Find the mean of the data in the dot plot below.
Annette [7]
Answer: 24 students

Explanation:

(20 + 23 + 25 + 28)/4
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3 years ago
A = 1/2h(c + d) for C.<br><br> Solve for c.
Vlada [557]

Answer:

am i right or wrong not sure.

Step-by-step explanation:

a=1/2hc+1/2hd

5 0
3 years ago
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
PilotLPTM [1.2K]

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!

3 0
3 years ago
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