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allochka39001 [22]
3 years ago
7

Bo paid bills with 1/2 of his paycheck snd put 1/5 of the remainder in savings. The rest of his paycheck he divided equally amon

g the college accounts of his 3 children.
A. what fraction of his paycheck went into each childs account?
B. If Bo deposited $400 in each childs account, how much money was in Bo's original paycheck?
Mathematics
1 answer:
kati45 [8]3 years ago
7 0
1/5 - 1/2 = 3/10
1/10 + 1/10 + 1/10 = 3/10

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There are many ways to measure the reading ability of children. Research designed to improve reading performance is dependent on
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Answer:

A 90% confidence interval to estimate the mean DRP score in Henrico County Schools  is =32±2.7280

i.e. C.I.  [29.3,34.7]

Hence the results are not significant.

Step-by-step explanation:

Given:

Total no of students =44

True mean =32

Standard deviation=11

And all 4 students score also given.

To Find:

90 % confidence interval and conclusion on it.

Solution:

Here for C.I. we required mean, standard deviation and no of students.

So mean =32,S.D.=11 and n=44

Therefore , For 90 % interval Zscore is Z=1.645

C.I.=mean±Z[Standard  deviation/Sqrt(n)]

=32±1.645[11/Sqrt(44)]

=32±1.645[1.6583]

=32±2.7280

i.e. C.I.  [29.3,34.7]

Now Here to commit on calculated interval we should check  value of Z-score

So we need to calculate the sample mean.

Sample mean =Sum of all values /Total no of values.

=[40+ 26 +39+ 14+ 42+ 18 +25+ 43+ 46 +27 +19+ 47+ 19 +26+ 35 +34+ 15+ 44 40+ 38+ 31+ 46 +52 +25+ 35 +35+ 33 +29 +34 +41 +49+ 28+ 42+ 47 +35 +48 +22 33+ 41 +51 +27 +14+ 54 +45]/44

=34.86

Z-score=(sample mean -true mean)/[standard deviation/Sqrt(n)]

=(34.86-32)/[11/Sqrt(44)]

Z=1.72465

For ,

P(Z≥1.72465)=0.08544

For 0.05 significance level  the results are not significant at  p<0.01.

Researcher claimed mean is not correct upto 0.05 significant level

i.e calculated mean and Sample mean are different.

Hence the results are not significant.

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Step-by-step explanation:

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