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Eddi Din [679]
3 years ago
5

Students who attend Anytown College pay either in-state or out-of-state tuition, depending on where they reside. The amount, I,

spent on in-state tuition and the amount, O, spent on out-of-state tuition for a randomly selected student both follow a Normal distribution.
The mean and standard deviation of the sum S = I + O of tuition for a randomly selected pair of in-state and out-of-state students are μS = $6,348.75 and σS = $1,508.48.

Use the z-table to answer the question.

What is the probability that the sum of a randomly selected pair of in- and out-of-state students has a tuition charge of more than $8,000?
0.0009
0.1368
0.8632
0.9991
Mathematics
1 answer:
notsponge [240]3 years ago
4 0

Answer:

The correct answer is 0.1368

Step-by-step explanation:

Got it right on Edge assignment

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The bus driver made 6 stops on his way into town. No passengers exited the bus. By stop 3, there were 9 passengers on the bus.
grigory [225]
It would take about 5 stops to get 13 people on the bus, but the real amount of people would be 15.
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On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas,
erastova [34]

Answer:

a

The distribution of  X is normal

b

P(X  <  1.4) = 0.12654

c

P(X  >  3.5) = 0.36051

Step-by-step explanation:

From the question we are told that

   The population mean is  \mu =  3 \  hours

   The standard deviation is  \sigma  =  1.4 \ hours

Generally given from the question that the amount of time spent alone by the population size is normally distributed then then the distribution of X (i.e the amount of time spent by the sample size (the one Mercurian)) will be normally distributed

  Generally the probability that the child spend less than one hour in a day is mathematically represented as

        P(X  <  1.4) =  P(\frac{X - \mu}{\sigma} < \frac{1.4 - \mu}{\sigma}   )

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

So

      P(X  <  1.4) =  P(Z < \frac{1.4 - 3.0}{1.4}   )

     P(X  <  1.4) =  P(Z < -1.1429  )

From the z-table the value of  

        P(Z < -1.1429  )=0.12654

So     P(X  <  1.4) = 0.12654

Generally the percentage of children that spends over 3.5 hours unsupervised is mathematically represented as

        P(X  >  3.5) =  P(\frac{X - \mu}{\sigma} > \frac{3.5 - \mu}{\sigma}   )

        P(X >  3.5) =  P(Z > \frac{3.5 - 3.0}{1.4}   )

        P(X >  3.5) =  P(Z > 0.3571  )

From the z-table the value of  

        P(Z >0.3571  )=0.36051

So     P(X  >  3.5) = 0.36051

5 0
3 years ago
A rectangular field is 125 yards long and the length of one diagonal of the field is 150 yards. What is the width of the field?
Artemon [7]
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6 0
3 years ago
The carnival committee has purchased 985 small prizes. The prizes are to be divided equally among the 29 game booths
Vladimir [108]

Answer:

1 The first digit of the quotient is 3

2. 34 prizes per booth

3.  19 prizes left over

Step-by-step explanation:

We need to divide 985 by 29


29 goes into 98   3 times

29*3 = 87  with 13 left over

135 divided by 29

29*4 =116  135-116 =19

985÷29 = 34  with 19 left over

7 0
3 years ago
a maple tree is 4.2 meters tall. A pine tree is 1.02 shorter that the maple tree. An oak tree is 1.89 taller than the pine tree
torisob [31]

Answer:

7.782352941

Step-by-step explanation:

4.2 divided by 1.02 the multiply the out come of that by 1.89 to get 7.782352941

5 0
4 years ago
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