Step-by-step explanation:
length = (2x + 3) units
width = (x - 7) units
Area = 2( length × width)
Area = 2( (2x + 3) × (x - 7) )
Area = 2( 2x² - 14x + 3x - 21)
Area = 2 ( 2x² -11x - 21)
Area = 4x² - 22x - 42
<em>Answer:</em>
<em>r = -</em>
<em />
<em>Step-by-step explanation:</em>
<em>Rewrite the equation as </em>
<em> = m</em>
<em>Remove the radical on the left side of the equation by squaring both sides of the equation.</em>
<em>(</em>
<em> = m^2</em>
<em>Then, you simplify each of the equation. </em>
<em>Rewrite: (</em>
<em> as </em>
<em />
<em>Remove any parentheses if needed.</em>
<em>Solve for r. </em>
<em>Multiply each term by r and simplify."</em>
<em>Multiply both sides of the equation by 5.</em>
<em>6a+r= m^2r⋅(5)</em>
<em>Remove parentheses.</em>
<em>Move 5 to the left of (m
^2) r
</em>
<em>6a+r=5m^2)r</em>
<em>Subtract 5m^2)r from both sides of the equation.</em>
<em>6a+r-5m^2)r=0</em>
<em>Subtract 6a from both sides of the equation.</em>
<em>r-5m^2)r=-6a</em>
<em>Factor r out of r-5m^2)r </em>
<em>r(1-5m^2)=-6a</em>
Divide each term by 1-5m^2 and simplify.
r = - 
There you go, hope this helps!
Answer:
6.18% of the class has an exam score of A- or higher.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:

What percentage of the class has an exam score of A- or higher (defined as at least 90)?
This is 1 subtracted by the pvalue of Z when X = 90. So



has a pvalue of 0.9382
1 - 0.9382 = 0.0618
6.18% of the class has an exam score of A- or higher.