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cricket20 [7]
3 years ago
15

quentin recorded the weather for the past 100 days. it rained on 28 of the days. which fraction and decimal represent how many d

ays it rained?
Mathematics
1 answer:
borishaifa [10]3 years ago
5 0
0.28 and 28/100(or 7/25)
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Peter is twice the age of Sara. If their total age is 27 years. how old is Sara?
Xelga [282]

Answer:

Just do 27x2 bro simple as that

Step-by-step explanation:

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Which equation is equivalent to
Aleksandr [31]

Answer:

(4x^2 +9)(4x^2-9) =0

Step-by-step explanation:

16x^4 - 18 = 0

Rewriting

(4x^2)^2 - 9^2 =0

We notice that this is the difference of squares

a^2 - b^2 = (a-b)(a+b)

(4x^2 -9)(4x^2+9) =0

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A merchant has 120 litres and 180 litres of two kinds of oil. He wants to sell the oil by filing the two kinds of equal volumes.
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HCF of 120 and 180 is 60 and hence a measuring tin of 60 litre volume should be bought.

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3 years ago
There are many ways to measure the reading ability of children. Research designed to improve reading performance is dependent on
ELEN [110]

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.

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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

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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.

6 0
2 years ago
ILL GIVE BRAINLIEST!!!<br><br> No need for explanation, just an answer is fine. Thank you.
Naddika [18.5K]

Answer:

10

Step-by-step explanation:

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2 years ago
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