Y = x/4
x = 4y
domain = {4(28), 4(30), 4(32), 4(34), 4(36)} = {112, 120, 128, 136, 144}
<span><span> y2(q-4)-c(q-4)</span> </span>Final result :<span> (q - 4) • (y2 - c)
</span>
Step by step solution :<span>Step 1 :</span><span>Equation at the end of step 1 :</span><span><span> ((y2) • (q - 4)) - c • (q - 4)
</span><span> Step 2 :</span></span><span>Equation at the end of step 2 :</span><span> y2 • (q - 4) - c • (q - 4)
</span><span>Step 3 :</span>Pulling out like terms :
<span> 3.1 </span> Pull out q-4
After pulling out, we are left with :
(q-4) • (<span> y2</span> * 1 +( c * (-1) ))
Trying to factor as a Difference of Squares :
<span> 3.2 </span> Factoring: <span> y2-c</span>
Theory : A difference of two perfect squares, <span> A2 - B2 </span>can be factored into <span> (A+B) • (A-B)
</span>Proof :<span> (A+B) • (A-B) =
A2 - AB + BA - B2 =
A2 <span>- AB + AB </span>- B2 =
<span> A2 - B2</span>
</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication.
Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.
Check : <span> y2 </span>is the square of <span> y1 </span>
Check :<span> <span> c1 </span> is not a square !!
</span>Ruling : Binomial can not be factored as the difference of two perfect squares
Final result :<span> (q - 4) • (y2 - c)
</span><span>
</span>
There would be an open circle on positive 12 with the arrow moving to the right.
Answer:


And we can find the limits in order to consider values as significantly low and high like this:


Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Solution to the problem
For this case we can consider a value to be significantly low if we have that the z score is lower or equal to - 2 and we can consider a value to be significantly high if its z score is higher tor equal to 2.
For this case we have the mean and the deviation given:


And we can find the limits in order to consider values as significantly low and high like this:


Answer:
Coordinates: (0,0) ; (1,5) ; (2,10)
Step-by-step explanation:
x | y
0 0
1 5
2 10
Coordinates: (0,0) ; (1,5) ; (2,10)
This should be a straight line that is going diagonally with a positive slope.