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Xelga [282]
2 years ago
5

for each x-y table given,copy the table, find the pattern and fill in the missing entries. Then write the rule for the pattern i

n words.​

Mathematics
1 answer:
mash [69]2 years ago
8 0

Answer:

  see the attached for the table and rules

Step-by-step explanation:

a) A graph of the given points shows they lie on the same line, one with a slope of 3 and a y-intercept of 2. Thus the rule is ...

  y = 3x +2

see the attachment for table values

__

b) The ratios of given points are all the same: y/x = 5/2, so that is the constant of proportionality:

  y = (5/2)x

see the attachment for table values

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An experiment on memory was performed, in which 16 subjects were randomly assigned to one of two groups, called "Sentences" or "
FromTheMoon [43]

Answer:

There is no significant difference between the averages.

Step-by-step explanation:

Let's call

\large X_{sentences} the mean of the “sentences” group

\large S_{sentences} the standard deviation of the “sentences” group

\large X_{intentional} the mean of the “intentional” group

\large S_{intentional} the standard deviation of the “intentional” group

Then, we can calculate by using the computer

\large X_{sentences}=28.75  

\large S_{sentences}=3.53553

\large X_{intentional}=31.625

\large S_{intentional}=1.40788

\large X_{sentences}-X_{intentional}=28.75-31.625=-2.875

The <em>standard error of the difference (of the means)</em> for a sample of size 8 is calculated with the formula

\large \sqrt{(S_{sentences})^2/8+(S_{intentional})^2/8}

So, the standard error of the difference is

\large \sqrt{(3.53553)^2/8+(1.40788)^2/8}=1.34546

<em>In order to see if there is a significant difference in the averages of the two groups, we compute the interval of confidence of  95% for the difference of the means corresponding to a level of significance of 0.05 (5%). </em>

<em>If this interval contains the zero, we can say there is no significant difference. </em>

<em>Since the sample size is small, we had better use the Student's t-distribution with 7 degrees of freedom (sample size-1), which is an approximation to the normal distribution N(0;1) for small samples. </em>

We get the \large t_{0.05} which is a value of t such that the area under the Student's t distribution  outside the interval \large [-t_{0.05}, +t_{0.05}] is less than 0.05.

That value can be obtained either by using a table or the computer and is found to be

\large t_{0.05}=2.365

Now we can compute our confidence interval

\large (X_{sentences}-X_{intentional}) \pm t_{0.05}*(standard \;error)=-2.875\pm 2.365*1.34546

and the confidence interval is

[-6.057, 0.307]

Since the interval does contain the zero, we can say there is no significant difference in these samples.

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docker41 [41]

Answer:

5

Step-by-step explanation:

7 0
1 year ago
Simplify 2n+5.5-0.9n-8+4.5p
Anettt [7]
The answer is 
1.1n+4.5p-2.5

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If a line crosses the yaxis at (0, 1) and has a slope of 4/5 , what is the equation of the line?
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Answer:

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Step-by-step explanation:

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