You can try to show this by induction:
• According to the given closed form, we have , which agrees with the initial value <em>S</em>₁ = 1.
• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume
and
We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or
From the given recurrence, we know
so that
which is what we needed. QED
Answer:
Anne is the best hitter as she has the best hitting average of 0.406
Step-by-step explanation:
Anne hits 65 bases out of 160 at - bats.
Beth's betting average = 0.399
Collin's hit = 40%
Now, we need to convert all the three hitting rates into same parameters to compare them with each other.
Average hit rate of Anne =
⇒The betting average of Anne = 0.406
Now, The hit - rate of Collin = 40%
Converting the percentage rate into decimal, we get the hit rate of Collin
=
Hence, we get that Hitting Average of :
<u>Anne = 0.406</u>, Beth = 0.3999 and Collin = 0.400
Hence, from the above data Anne is the best hitter as she has the best hitting average of 0.406.
Answer:
Step-by-step explanation:
y > 3x + 10
y < -3x - 1
<h3> Part A:</h3>You'd graph the given systems of linear inequalities the same way as you graph the linear equations.
To graph y > 3x + 10, plot the y-intercept, (0, 10), then use the slope, m = 3 (rise 3, run 1), to plot other points on the graph. Use a dashed line (because of the ">" symbol).
Follow the same steps for the other linear inequality. Plot the y-intercept, (0, -1), then use the slope, m = -3 (down 3, run 1) to plot other points. Use a dashed line (because of the "<" symbol).
Pick a test point on either side of the boundary line and plug it into the original problem. This will help determine which side of the boundary line is the solution. Plug in a test point that is not on the boundary line.
Use the point of origin, (0, 0) as the test point. Plug in these values into the given systems of linear inequalities to see whether it will provide a true statement.
y > 3x + 10
0 > 3(0) + 10
0 > 0 + 10
0 > 10 (False statement).
y < -3x - 1
0 < -3(0) - 1
0 < 0 - 1
0 < -1 (false statement).
Since the point of origin provided a false statement to the given systems of linear inequalities, you must shade the half-plane region where it doesn't contain the test point.
<h3>Part B: </h3>
You'll do the same process as what I've done for the test point. Plug in the values of (8, 10) into the given systems of linear inequalities. If it provides a false statement, then it means that it is not a solution to the system.
y > 3x + 10
10 > 3(8) + 10
10 > 24 + 10
10 > 34 (False statement).
y < -3x - 1
10 < -3(8) - 1
10 < -24 - 1
10 < -25 (false statement).
Therefore, (8, 10) is not a solution to the system.
