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sashaice [31]
3 years ago
6

A Kamal has 6 cans of regular soda and 15 cans of diet soda. He wants to create some identical refreshment tables that will oper

ate during the American football game. He also doesn't want to have any sodas left over. What is the greatest number of refreshment tables that Kamal can stock? Pls I need help. TwT
Mathematics
1 answer:
N76 [4]3 years ago
4 0

Answer

15 + 6 = 21 15 - 6 = 9 is the answer

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Answer:

The correct answer is 14 not 17 or 7

Step-by-step explanation:

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A duplex generates $1,400 rent for each of its two units per month. The property has been recently appraised for $350,000. What
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Jill swam 200 meters in 2 minutes 42 seconds. If each lap is 50 meters long, which time is her estimated time, in seconds, per l
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Answer

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Step-by-step explanation:

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3 years ago
Read 2 more answers
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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4 years ago
Write an equation of a line in slope-intercept form when the slope is 2 and the y-intercept is -3
kupik [55]

Answer:

2x-3

Step-by-step explanation:


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