QuestionIn 2003, card usage was $41\%$ and is projected to be $61\%$ in 2011.
(a) Find a linear function $P(x)$ for percentage over years. (b) Estimate when percentage was between $46\%$ and $56\%$ .

Studdy Solution

STEP 1

Assumptions1. The percentage of purchases paid by credit and debit cards in2003 is41% . The projected percentage in2011 is61%
3. The relationship between the year and the percentage of purchases paid by credit and debit cards is linear4. $x$ is the year and $(x)$ is the percentage

STEP 2

We need to find a linear function that models the data. A linear function has the form $(x) = mx + b$ , where $m$ is the slope and $b$ is the y-intercept.

STEP 3

First, we need to find the slope $m$ . The slope is the change in $(x)$ divided by the change in $x$ . We can calculate it as follows $m = \frac{(x2) -(x1)}{x2 - x1}$

STEP 4

Now, plug in the given values for $(x1)$ , $(x2)$ , $x1$ , and $x2$ to calculate the slope. Here, $x1$ is2003, $(x1)$ is41%, $x2$ is2011, and $(x2)$ is61%.
$m = \frac{61\% -41\%}{2011 -2003}$

STEP 5

Calculate the slope.
$m = \frac{20\%}{8} =2.5\%$

STEP 6

Now that we have the slope, we can find the y-intercept $b$ . We can calculate it by rearranging the linear function to solve for $b$ $b =(x) - mx$

STEP 7

Plug in the values for $(x)$ , $m$ , and $x$ to calculate the y-intercept. Here, we can use the data from2003, so $(x)$ is41% and $x$ is2003.
$b =41\% -2.5\% \times2003$

STEP 8

Calculate the y-intercept.
$b =41\% -2.5\% \times2003 = -4982.5\%$

STEP 9

Now we have the slope and the y-intercept, so we can write the linear function that models the data $(x) =2.5\%x -4982.5\%$

STEP 10

To estimate when the percentage was between46% and56%, we can set $(x)$ equal to these values and solve for $x$ .
First, set $(x)$ equal to46% $46\% =2.5\%x -4982.5\%$

STEP 11

Rearrange the equation to solve for $x$ $x = \frac{46\% +498.5\%}{.5\%}$

STEP 12

Calculate $x$ .
$x = \frac{46\% +4982.5\%}{2.5\%} =2005.8$

STEP 13

Now, set $(x)$ equal to56% $56\% =2.5\%x -4982.5\%$

STEP 14

Rearrange the equation to solve for $x$ $x = \frac{56\% +4982.\%}{2.\%}$

STEP 15

Calculate $x$ .
$x = \frac{56\% +4982.5\%}{2.5\%} =2007.4$ So, the percentage of purchases paid by credit and debit cards was between46% and56% between the years2005.8 (around the end of2005) and2007.4 (around the middle of2007).

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QuestionIn 2003, card usage was $41\%$ and is projected to be $61\%$ in 2011.
(a) Find a linear function $P(x)$ for percentage over years. (b) Estimate when percentage was between $46\%$ and $56\%$ .

Studdy Solution

STEP 1

Assumptions1. The percentage of purchases paid by credit and debit cards in2003 is41% . The projected percentage in2011 is61%
3. The relationship between the year and the percentage of purchases paid by credit and debit cards is linear4. $x$ is the year and $(x)$ is the percentage

STEP 2

We need to find a linear function that models the data. A linear function has the form $(x) = mx + b$ , where $m$ is the slope and $b$ is the y-intercept.

STEP 3

First, we need to find the slope $m$ . The slope is the change in $(x)$ divided by the change in $x$ . We can calculate it as follows $m = \frac{(x2) -(x1)}{x2 - x1}$

STEP 4

Now, plug in the given values for $(x1)$ , $(x2)$ , $x1$ , and $x2$ to calculate the slope. Here, $x1$ is2003, $(x1)$ is41%, $x2$ is2011, and $(x2)$ is61%.
$m = \frac{61\% -41\%}{2011 -2003}$

STEP 5

Calculate the slope.
$m = \frac{20\%}{8} =2.5\%$

STEP 6

Now that we have the slope, we can find the y-intercept $b$ . We can calculate it by rearranging the linear function to solve for $b$ $b =(x) - mx$

STEP 7

Plug in the values for $(x)$ , $m$ , and $x$ to calculate the y-intercept. Here, we can use the data from2003, so $(x)$ is41% and $x$ is2003.
$b =41\% -2.5\% \times2003$

STEP 8

Calculate the y-intercept.
$b =41\% -2.5\% \times2003 = -4982.5\%$

STEP 9

Now we have the slope and the y-intercept, so we can write the linear function that models the data $(x) =2.5\%x -4982.5\%$

STEP 10

To estimate when the percentage was between46% and56%, we can set $(x)$ equal to these values and solve for $x$ .
First, set $(x)$ equal to46% $46\% =2.5\%x -4982.5\%$

STEP 11

Rearrange the equation to solve for $x$ $x = \frac{46\% +498.5\%}{.5\%}$

STEP 12

Calculate $x$ .
$x = \frac{46\% +4982.5\%}{2.5\%} =2005.8$

STEP 13

Now, set $(x)$ equal to56% $56\% =2.5\%x -4982.5\%$

STEP 14

Rearrange the equation to solve for $x$ $x = \frac{56\% +4982.\%}{2.\%}$

STEP 15

Calculate $x$ .
$x = \frac{56\% +4982.5\%}{2.5\%} =2007.4$ So, the percentage of purchases paid by credit and debit cards was between46% and56% between the years2005.8 (around the end of2005) and2007.4 (around the middle of2007).

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QuestionIn 2003, card usage was 41%41\%41% and is projected to be 61%61\%61% in 2011. (a) Find a linear function P(x)P(x)P(x) for percentage over years. (b) Estimate when percentage was between 46%46\%46% and 56%56\%56%.

Studdy Solution

STEP 1

Assumptions1. The percentage of purchases paid by credit and debit cards in2003 is41% . The projected percentage in2011 is61%3. The relationship between the year and the percentage of purchases paid by credit and debit cards is linear4. xxx is the year and (x)(x)(x) is the percentage

STEP 2

We need to find a linear function that models the data. A linear function has the form (x)=mx+b(x) = mx + b(x)=mx+b, where mmm is the slope and bbb is the y-intercept.

STEP 3

First, we need to find the slope mmm. The slope is the change in (x)(x)(x) divided by the change in xxx. We can calculate it as followsm=(x2)−(x1)x2−x1m = \frac{(x2) -(x1)}{x2 - x1}m=x2−x1(x2)−(x1)​

STEP 4

Now, plug in the given values for (x1)(x1)(x1), (x2)(x2)(x2), x1x1x1, and x2x2x2 to calculate the slope. Here, x1x1x1 is2003, (x1)(x1)(x1) is41%, x2x2x2 is2011, and (x2)(x2)(x2) is61%.m=61%−41%2011−2003m = \frac{61\% -41\%}{2011 -2003}m=2011−200361%−41%​

STEP 5

Calculate the slope.m=20%8=2.5%m = \frac{20\%}{8} =2.5\%m=820%​=2.5%

STEP 6

Now that we have the slope, we can find the y-intercept bbb. We can calculate it by rearranging the linear function to solve for bbbb=(x)−mxb =(x) - mxb=(x)−mx

STEP 7

Plug in the values for (x)(x)(x), mmm, and xxx to calculate the y-intercept. Here, we can use the data from2003, so (x)(x)(x) is41% and xxx is2003.b=41%−2.5%×2003b =41\% -2.5\% \times2003b=41%−2.5%×2003

STEP 8

Calculate the y-intercept.b=41%−2.5%×2003=−4982.5%b =41\% -2.5\% \times2003 = -4982.5\%b=41%−2.5%×2003=−4982.5%

STEP 9

Now we have the slope and the y-intercept, so we can write the linear function that models the data(x)=2.5%x−4982.5%(x) =2.5\%x -4982.5\%(x)=2.5%x−4982.5%

STEP 10

To estimate when the percentage was between46% and56%, we can set (x)(x)(x) equal to these values and solve for xxx.First, set (x)(x)(x) equal to46%46%=2.5%x−4982.5%46\% =2.5\%x -4982.5\%46%=2.5%x−4982.5%

STEP 11

Rearrange the equation to solve for xxxx=46%+498.5%.5%x = \frac{46\% +498.5\%}{.5\%}x=.5%46%+498.5%​

STEP 12

Calculate xxx.x=46%+4982.5%2.5%=2005.8x = \frac{46\% +4982.5\%}{2.5\%} =2005.8x=2.5%46%+4982.5%​=2005.8

STEP 13

Now, set (x)(x)(x) equal to56%56%=2.5%x−4982.5%56\% =2.5\%x -4982.5\%56%=2.5%x−4982.5%

STEP 14

Rearrange the equation to solve for xxxx=56%+4982.%2.%x = \frac{56\% +4982.\%}{2.\%}x=2.%56%+4982.%​

STEP 15

Calculate xxx.x=56%+4982.5%2.5%=2007.4x = \frac{56\% +4982.5\%}{2.5\%} =2007.4x=2.5%56%+4982.5%​=2007.4So, the percentage of purchases paid by credit and debit cards was between46% and56% between the years2005.8 (around the end of2005) and2007.4 (around the middle of2007).

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QuestionIn 2003, card usage was $41\%$ and is projected to be $61\%$ in 2011.
(a) Find a linear function $P(x)$ for percentage over years. (b) Estimate when percentage was between $46\%$ and $56\%$ .

Assumptions1. The percentage of purchases paid by credit and debit cards in2003 is41% . The projected percentage in2011 is61%
3. The relationship between the year and the percentage of purchases paid by credit and debit cards is linear4. $x$ is the year and $(x)$ is the percentage

We need to find a linear function that models the data. A linear function has the form $(x) = mx + b$ , where $m$ is the slope and $b$ is the y-intercept.

First, we need to find the slope $m$ . The slope is the change in $(x)$ divided by the change in $x$ . We can calculate it as follows $m = \frac{(x2) -(x1)}{x2 - x1}$

Now, plug in the given values for $(x1)$ , $(x2)$ , $x1$ , and $x2$ to calculate the slope. Here, $x1$ is2003, $(x1)$ is41%, $x2$ is2011, and $(x2)$ is61%.
$m = \frac{61\% -41\%}{2011 -2003}$

Calculate the slope.
$m = \frac{20\%}{8} =2.5\%$

Now that we have the slope, we can find the y-intercept $b$ . We can calculate it by rearranging the linear function to solve for $b$ $b =(x) - mx$

Plug in the values for $(x)$ , $m$ , and $x$ to calculate the y-intercept. Here, we can use the data from2003, so $(x)$ is41% and $x$ is2003.
$b =41\% -2.5\% \times2003$

Calculate the y-intercept.
$b =41\% -2.5\% \times2003 = -4982.5\%$

Now we have the slope and the y-intercept, so we can write the linear function that models the data $(x) =2.5\%x -4982.5\%$

To estimate when the percentage was between46% and56%, we can set $(x)$ equal to these values and solve for $x$ .
First, set $(x)$ equal to46% $46\% =2.5\%x -4982.5\%$

Rearrange the equation to solve for $x$ $x = \frac{46\% +498.5\%}{.5\%}$

Calculate $x$ .
$x = \frac{46\% +4982.5\%}{2.5\%} =2005.8$

Now, set $(x)$ equal to56% $56\% =2.5\%x -4982.5\%$

Rearrange the equation to solve for $x$ $x = \frac{56\% +4982.\%}{2.\%}$

Calculate $x$ .
$x = \frac{56\% +4982.5\%}{2.5\%} =2007.4$ So, the percentage of purchases paid by credit and debit cards was between46% and56% between the years2005.8 (around the end of2005) and2007.4 (around the middle of2007).