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trapecia [35]
3 years ago
15

Identify two segments that are marked congruent to each other on the diagram below. ANSWER CORRECTLY !!!!!!!!!! WILL

Mathematics
1 answer:
Molodets [167]3 years ago
6 0

Answer:

IM and JM

Step-by-step explanation:

They both have congruency marks that correspond to each other, indicated by the three short lines in the middle of each.

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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.

6 0
4 years ago
What is the sum ? (2x^2-4x+1)+(5x+x^2-1)
Ipatiy [6.2K]

Hello There!

<u>The answer is...</u>

<u />

<u />3x^2+x<u />

<u />

HERE'S A EXPLANATION!

2x^2-4x+1+5x+x^2-1

=2x^2+-4x+1+5x+x^2+-1

<u>Combine these like terms!</u>

<u></u>

<u></u>=2x^2+-4x+1+5x+x^2+-1<u></u>

<u></u>

<u></u>=(2x^2+x^2)+(-4x+5x)+(1+-1)<u></u>

<u></u>

<u></u>=3x^2+x<u></u>

<u></u>

<u>ANSWER!</u>

<u />

<u />=3x^2+x<u />

<u />

Hopefully, this helps you!!

AnimeVines

6 0
3 years ago
Help me pls (: l: ):
densk [106]

Fraction: 5/10

Demical: 0.5

Percent: 50%

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3 years ago
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How many 1/2s are in 3 1/2?
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7 halves are in 3.5
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3 years ago
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Find the least common multiple of x3 - x2 + x - 1 and x2 - 1. Write the answer in factored form. A. (x + 1)2(x - 1) B. (x + 1)(x
WINSTONCH [101]
Hello,
Answer B

x^3-x^2+x-1=(x-1)(x^2+1)\\&#10;x^2-1=(x+1)(x-1)\\&#10;lcm=(x-1)(x+1)(x^2+1)\\&#10;
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4 years ago
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