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Minchanka [31]
3 years ago
7

(2 5/6)(6)+(-1 1/2)(1.5)-(1/16)

Mathematics
1 answer:
Jet001 [13]3 years ago
4 0
Now I’m ngl my math might be wrong but I ended up with 1131/48
As a decimal I got 23.6
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The distance between point (-6,-4) and (-6,5)
trasher [3.6K]

Answer:

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Step-by-step explanation:

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2 years ago
If the actual distribution of tusk lengths does not match the ones predicted with the H-W equation, it may indicate that natural
Anna [14]

Answer:

See explanation below

Step-by-step explanation:

It depends on what null hypothesis is under consideration.

One of the most common null hypothesis that are subject of study in a given statistical model is <em>the mean</em> predicted by the model.

In this case, the scientist probably observed that the mean of tusk lengths she obtained in a sample did not match the one predicted with the H-W equation.

So, she decided to perform a statistical study by collecting random samples and measuring the tusk lengths to determine a new possible mean and contrast it against the one predicted by the H-W equation.

<em>Let's call M the mean predicted by the H-W equation, and S the mean obtained by the scientist. </em>

If M different of S and the p-value is 0.021, that means that <em>there is at most 2.1% of probability that the difference between M and S could be due to a random sampling error. </em>

It should be kept in mind that the p-value does not represent the probability that the scientist is wrong.

7 0
3 years ago
Graph the line with slope -3/4, passing through the points −3−3<br> .
snow_tiger [21]
Hello:
equation is the line is : y = ax+b        a is a slope
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y = (-3/4)x-(21/4)

</span>We draw the <span>Graph by two points : A(-3;-3)  B(0;-21/4)</span>

5 0
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Sergeeva-Olga [200]

f(\stackrel{x_1}{-4})=\stackrel{y_1}{0}\qquad f(\stackrel{x_2}{0})=\stackrel{y_2}{0}~\hfill (\stackrel{x_1}{-4}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-4)}}}\implies \cfrac{0}{0+4}\implies \cfrac{0}{4}\implies 0

\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{0}=\stackrel{m}{0}(x-\stackrel{x_1}{(-4)})\implies \boxed{y = 0}

6 0
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Which of the following is true of an ellipse with foci in the same location as the center?
uranmaximum [27]

Answer:c

Step-by-step explanation:

c

3 0
3 years ago
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