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2 years ago

Describe how you regroup when you find the sum of 64+43

1 answer:

5 0

Here, regrouping is basically carrying.

64+43 shown vertically would be:

64
+43
-------
107
4+3 is 7, so seven is in the ones place, but that's not the point.
60+40 is 100, so you regroup by carrying the one to the hundreds place.

Hope I helped!

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If Serena selects 4 samples from the table, which row

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Answer:1: Row 3

2: Row 2

3:6 and 10

Step-by-step explanation:

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1 year ago

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As a laboratory assistant, you measure chemicals using the metric system. For your current research, you need to measure out 45

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Compute E(X) for the following random variable X : X=Number of tosses until all 10 numbers are seen (including the last toss) by

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Answer:

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1 year ago

TWENTY POINTS!!! Baxter is thinking about buying a car for $18,500. The table below shows the projected value of two different c

Anika [276]

Part A: Linear function

explanation:
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1 18,500 18,500
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2 years ago

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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 standard error of the difference (of the means) 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

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

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%).

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

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.

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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2 years ago