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nekit [7.7K]
3 years ago
9

How many reasons must your thesis contain and which reason is first

Mathematics
1 answer:
finlep [7]3 years ago
5 0
<span><span>The more sources the better
Every researcher aims to have a standardized and uniform study when a study will be replicated or repeated for various reasons.

</span>The number of accounts and evidential support of your thesis must be consistent and be sufficient and succinct to the topic of study. These reasons and evidential support should claim proximate with the pronounced experimentation and variables which will be studied. In sense, the more accounts and sources the better is your study, the greater it will stand out and present itself. This is important for every thesis paper or dissertation since these accounts provides and explicitly discusses the studies that have been undertaken before which can greatly influence the current study's credibility and reliability. </span>

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The annual precipitation amounts in a certain mountain range are normally distributed with a mean of 104 inches, and a standard
dsp73

Answer:

91.92% probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 104, \sigma = 14, n = 49, s = \frac{14}{\sqrt{49}} = 2

What is the probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

This is the pvalue of Z when X = 106.8. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{106.8 - 104}{2}

Z = 1.4

Z = 1.4 has a pvalue of 0.9192

0.9192 = 91.92% probability that the mean annual precipitation during 49 randomly picked years will be less than 106.8 inches

8 0
3 years ago
After sailing 12 mi, a sailor changed direction and increased the boat's speed by 4 mph. an additional 18 mi was sailed at the i
inessss [21]
Let the speed for the first 12 mi be x mi/h, the speed for 18 mi was (x+4) mi/h
thus given that
time=distance/speed 
the average time will be:
3=(12+18)/(x+x+4)
3=30/(2x+4)
solving for x we get
3(2x+4)=30
6x+6=30
6x=24
x=4 mi/hr
Answer: 12 mi/hr
8 0
3 years ago
A researcher computes the following 2 x 3 between-subjects ANOVA, in which 11 participants were observed in each group. Which ef
pishuonlain [190]

Answer:

D. There were no significant effects.

Step-by-step explanation:

The table below shows the representation of the significance level using the two-way between subjects ANOVA.

Source of Variation            SS          df          MS          F                P-value

Factor A                             10          1              10           0.21           0.660

Factor B                             50         2             25          0.52          0.6235  

     

A × B                                  40          2             20         0.42           0.6783

Error                                   240        5             48           -               -

Total                                   340       10             -             -                  -

From the table above , the SS(B) is determined as follows:

SS(B) = SS(Total)-SS(Error-(A×B)-A)

         =  340-(240-40-10)

         = 50

A researcher computes the following 2 x 3 between-subjects ANOVA;

k=2

n=3

N(total) = no of participants observed in each group =11

df for Factor A= (k-1)

=(2-1)

=1

df for Factor B = (n-1)

=(3-1)

=2

df for A × B    

 = 2 × 1

 = 2

df  factor for total

=(N-1)

=11-1

=10

MS = SS/df

Thus, from the table, the P-Value for all data is greater than 0.05, therefore we fail to reject H₀.

4 0
3 years ago
Solve (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0<br> x = 5<br> x = 6<br> x = 7<br> x = 8
Anna11 [10]

Answer:

x = 6

Step-by-step explanation:

3 0
3 years ago
HELLP HEELEP SOS EMERGENCY
icang [17]

Answer:

15 in.

Step-by-step explanation:

P = 2(L + W)

36 = 2(L + 3)

18 = L + 3

L = 15

3 0
2 years ago
Read 2 more answers
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