What is the slope of the line that passes through the points (26, 7) and (-39, 12)
1 answer:
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Answer:
Step-by-step explanation:
We have the function:
And we want to find x=d for which the minimum/maximum value will occur.
Notice that our function is a quadratic in factored form.
Remember that the minimum/maximum value always occurs at the vertex point.
And remember that the x-coordinate of the vertex is the axis of symmetry.
Since a quadratic is always symmetrical on both sides of its axis of symmetry, a quadratic’s axis of symmetry is the average of the two roots/zeros of the quadratic.
Therefore, the value x=d such that it produces the minimum/maximum value is the average of the two roots.
Our factors are <em>(x-2) </em>and <em>(x-1)</em>.
Therefore, our roots/zeros are <em>x=1, 2</em>.
So, the average of them are:
Therefore, regardless of the value of <em>a</em>, the minimum/maximum value will occur at <em>x=d=1.5</em>.
Alternative Method:
Of course, we can also expand to confirm our answer. So:
The x-coordinate of the vertex is still going to be the place where the minimum/maximum is going to occur.
And the formula for the vertex is:
So, we will substitute <em>-3a</em> for <em>b</em> and <em>a</em> for <em>a</em>. This yields:
Confirming our answer.
Correct Ans:
Option A. 0.0100
Solution:
We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:
So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.
Option A. 0.0100
Solution:
We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.
First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:
So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.
Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.