Use slope formula: y2-y1/x2-x1
XZ: -5/9
YZ: 2/7
XY: -7/2
XYZ is a right triangle because two of these slopes have a product of -1.
(YZ and XY does). We are all busy these days.
The correct answer is H correct me if I’m wrong but I looked it up also branilest plz
Answer:
Dan is correct
Step-by-step explanation:
From the question, we have:

Required
The average
The average hour is calculated using:

So, we have:



<em>Dan calculated the mean as 3; while Bret calculated the mean as 4.</em>
<em>Hence, we can conclude that Dan s correct</em>
Answer:

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution
and for this case we know that the distribution is given by:

And the standard error would be:

And replacing we got:

Step-by-step explanation:
For this case we know the population deviation given by:

And we have a sample size of n =81. We want to estimate the standard error of the sampling distribution
and for this case we know that the distribution is given by:

And the standard error would be:

And replacing we got:

Answer:

Step-by-step explanation:

<u>Multiply both sides of the equation by 6:</u>

<u>Subtract 12 from both sides:</u>

<u>Subtract 12 from 96:</u>

<u>_________________________________</u>