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sergiy2304 [10]
3 years ago
6

The average American generates 4.4 lbs of trash every day. We circulated fliers that listed tips on how to reduce wastefulness i

n three separate neighborhoods. For the next week, we measured the amount of trash each person produced each day in those neighborhoods. There were 625 people total in our study. The mean trash per person was 4.3 lbs with a standard deviation of 1
Determine your sample's score on the comparison distribution.
a) -2
b) -1
c) -1.5
d) -2.5
Mathematics
1 answer:
Hatshy [7]3 years ago
6 0

Answer:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z=\frac{4.3 -4.4}{\frac{1}{\sqrt{625}}}= -2.5

And the best option would be:

d) -2.5

Step-by-step explanation:

For this problem we know that the true mean of  trash every day is:

\mu =4.4

And from the info given we also know that:

\bar X=4.3 represent the sample mean

n=625 sample size selected

\sigma = 1 the population standard deviation assumed

If we want to find the z score for the person we can use the following formula:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z=\frac{4.3 -4.4}{\frac{1}{\sqrt{625}}}= -2.5

And the best option would be:

d) -2.5

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3 years ago
Find z1z2 if z1 = 2/5(cos63° + isin63°) and z2 = 10/3(cos117° + isin117°).
Airida [17]

Answer:

-10/3 is the answer

Step-by-step explanation:

Given are two complex numbers as

z1=\frac{2}{5} (cos63  + isin63)\\z2=\frac{10}{3} (cos 117 + isin117})

Recall Demoivre theorem as

(cosA+isinA)(cos A+isin B) = cos(A+B)+isin(A+B)

Hence here we have sum of angles  as

A+B = 63+117 =180

z1z2=\frac{2}{5} \frac{10}{3} (cos180+isin180)

=\frac{-10}{3}

Since sin180=0 and cos 180=-1

7 0
3 years ago
An Airliner has a capacity for 300 passengers. If the company overbook a flight with 320 passengers, What is the probability tha
TEA [102]

Answer:

The probability is   P(X  >300 ) = 0.97219

Step-by-step explanation:

From the question we are told that

 The capacity of  an Airliner  is  k =  300 passengers

 The sample size n =  320 passengers

  The probability the a randomly selected passenger shows up on to the airport

    p = 0.96

Generally the mean is mathematically represented as

    \mu  =  n*  p

  => \mu  =  320 *  0.96

    => \mu  = 307.2

Generally the standard deviation is  

    \sigma =  \sqrt{n *  p *  (1 -p ) }

=>  \sigma =  \sqrt{320  *  0.96 *  (1 -0.96 ) }

=> \sigma =3.50

Applying Normal approximation of binomial distribution

Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

  P(X  > k ) =  P( \frac{ X -\mu }{\sigma }  >  \frac{k - \mu}{\sigma } )

Here \frac{ X -\mu }{\sigma }  =Z (The \ standardized \  value \  of  \ X )

=>P(X  >300 ) =  P(Z >  \frac{300 - 307.2}{3.50} )

Now applying  continuity correction we have

    P(X  >300 ) =  P(Z >  \frac{[300+0.5] - 307.2}{3.50} )    

=>    P(X  >300 ) =  P(Z >  \frac{[300.5] - 307.2}{3.50} )

=>    P(X  >300 ) =  P(Z >  -1.914 )

From the z-table  

    P(Z >  -1.914 ) =  0.97219

So

    P(X  >300 ) = 0.97219

8 0
3 years ago
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