Percent increase from 2000 to 2001 is 12% and from 2001 to 2002 is 12.6%. Thus period of 2001 to 2002 has higher increase percentage.
<u>Solution:</u>
Given that,
Total circulation of local newspaper in 2000 = 3250
Total circulation of local newspaper in 2001 = 3640
Total circulation of local newspaper in 2002 = 4100
<em><u>Finding percent of increase in the newspaper’s circulation from 2000 to 2001:</u></em>
![\text { Percent increase from } 2000 \text { to } 2001=\frac{\text { change in circulation from 2001 - 2000 }}{\text { circulation in } 2000} \times 100](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Percent%20increase%20from%20%7D%202000%20%5Ctext%20%7B%20to%20%7D%202001%3D%5Cfrac%7B%5Ctext%20%7B%20change%20in%20circulation%20from%202001%20-%202000%20%7D%7D%7B%5Ctext%20%7B%20circulation%20in%20%7D%202000%7D%20%5Ctimes%20100)
![\begin{array}{l}{=\frac{3640-3250}{3250} \times 100} \\\\ {=\frac{390}{3250} \times 100=12 \%}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%3D%5Cfrac%7B3640-3250%7D%7B3250%7D%20%5Ctimes%20100%7D%20%5C%5C%5C%5C%20%7B%3D%5Cfrac%7B390%7D%7B3250%7D%20%5Ctimes%20100%3D12%20%5C%25%7D%5Cend%7Barray%7D)
<em><u>Finding percent of increase in the newspaper’s circulation from 2001 to 2002:</u></em>
![\begin{array}{l}{\text { Percent increase from } 2001 \text { to } 2002=\frac{\text { change in circulation }}{\text {circulation in } 2001} \times 100} \\\\ {=\frac{4100-3640}{3640} \times 100} \\\\ {=\frac{460}{3640} \times 100=12.6 \%}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7B%5Ctext%20%7B%20Percent%20increase%20from%20%7D%202001%20%5Ctext%20%7B%20to%20%7D%202002%3D%5Cfrac%7B%5Ctext%20%7B%20change%20in%20circulation%20%7D%7D%7B%5Ctext%20%7Bcirculation%20in%20%7D%202001%7D%20%5Ctimes%20100%7D%20%5C%5C%5C%5C%20%7B%3D%5Cfrac%7B4100-3640%7D%7B3640%7D%20%5Ctimes%20100%7D%20%5C%5C%5C%5C%20%7B%3D%5Cfrac%7B460%7D%7B3640%7D%20%5Ctimes%20100%3D12.6%20%5C%25%7D%5Cend%7Barray%7D)
As we can see 12% < 12.6%
So period of 2001 to 2002 has higher increase percentage.
Answer:
1. y = mx + b
Step-by-step explanation:
y = mx + b
2) to find the y intercept you need to see what cross | ( 8,5)
3) 5 = m(8) = b
4) your slope is negative form the why it pointing
5) in this case your slope is 1/3
6) so now you can do your first step now that you have your (y,x) and slope (your solving for (b)
7)
5 = 1/3 (8) + b
5 = 2.6 + b
5 = 2.6
-2.6
2.4 = b
Take the sqrt of both sides of <span>(x+2)^2=10:
x+2 = plus or minus sqrt(10)
Solve for x: x = -2 plus or minus sqrt(10)
Evaluate x: x = -2 + sqrt(10) AND x = -2 - sqrt(10) (answers)
</span>
Answer:
I believe the answer would be 32.64
Step-by-step explanation:
$24 x 36% = 8.64
8.64 + 24 = $32.64
Answer:
In order to have ran 33 miles, Bobby would have to attend <em>32 track practices.</em>
Step-by-step explanation:
Solving this problem entails of uncovering the amount of track practices Bobby must attend in order to have ran 33 miles. Start by reading the problem carefully to break down the information provided.
You can see that Bobby has already ran one mile on his own. This is important to remember for later. The problem also states that he expects to run one mile at every track practice.
Setting up an equation will help us solve. Here is how we could set up the equation:
(<em>amount of miles already ran</em> = 1) + (<em>number of track practices</em> = x) = (<em>total miles to run</em> = 33)
1 + x = 33
The equation is now in place. You can solve this, or isolate <em>'x',</em> by using the subtraction property of equality. This means we will subtract one from both sides of the equation, thus isolating the variable.
1 + x = 33
1 - 1 + x = 33 - 1
x = 32
The variable is the only term left on the left side of the equation. This means Bobby must attend track practice <em>32 times</em> in order to have ran 33 miles.