Ask question

Ask question

All categories

1 year ago

10

Leon has $250 in an account. Every 15 days, he withdraws $50. The amount of money remaining in his account is a function of the

amount of time that has passed. Which function models the amount of money remaining in Leon’s account?

1 answer:

Allisa [31]1 year ago

7 0

A step function, as it decreases once every 15 days and remains constant for periods in between.

You might be interested in

Lenovo uses the zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as follows:

Stella [2.4K]

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

$\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}$

The forcast for a period $F_{t+1}$ is given by the formular

$F_{t+1}=\alpha A_t+(1-\alpha)F_t$

where $A_t$ is the actual value for the preceding period and $F_t$ is the forcast for the preceding period.

Part 1A:
Given <span>α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804$

Therefore, the foreast for period 11 is $1.80

Part 1B:

</span>Given <span>α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Therefore, the foreast for period 12 is $1.78</span>

Part 2A:

Given <span>α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70

</span>
<span><span>Part 2B:

</span>Given <span>α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68

</span></span>
<span>Part 3A:

Given <span>α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65

</span>
<span><span>Part 3B:

</span>Given <span>α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

Therefore, the forecast for period 12 is $1.64

Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span></span></span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

$\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157

</span><span><span>Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = 
\frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = 
\frac{0.32}{3} \approx0.107$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107

</span></span>
<span><span>Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = 
\frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = 
\frac{29}{3} \approx0.097$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>

5 0

1 year ago

9x-2y=11 <br> 5x-2y=15 <br> what is the solution to the above system of equations<br> ASAP

natali 33 [55]

9x - 2y = 11 ... (i)

5x - 2y = 15 ... (ii)

Subtracting equation (ii) from (i) we get;

4x + 0 = -4

4x=-4 , x = -1

Replacing x = -1 in equation (i) we get;

9(-1) - 2y = 11

-9 - 2y = 11

-2y = 20

y = 20 ÷ -2 = -10

The solution to the system of equations is (-1,-10).

8 0

1 year ago

Read 2 more answers

The power generated by an electrical circuit (in watts) as a function of its current x xx (in amperes) is modeled by P ( x ) = −

marysya [2.9K]

Given:

The power generated by an electrical circuit (in watts) as a function of its current x (in amperes) is modeled by

$P(x)=-15x(x-8)$

To find:

The current which will produce the maximum power.

Solution:

We have,

$P(x)=-15x(x-8)$

$P(x)=-15x^2+120x$

Differentiate with respect to x.

$P'(x)=-15(2x)+120(1)$

$P'(x)=-30x+120$ ...(i)

To find the extreme point equate P'(x)=0.

$-30x+120=0$

$-30x=-120$

Divide both sides by -30.

$x=4$

Differentiate (i) with respect to x.

$P'(x)=-30(1)+(0)$

$P'(x)=-30$ (Maximum)

It means, the given function is maximum at x=4.

Therefore, the current of 4 amperes will produce the maximum power.

7 0

2 years ago

Relative to the graph of y=3sinx what is the shift of the graph of y=3sin(x+pi/3)

DochEvi [55]

The shifts of the sinus function can be described with the formula:
<span>a<span>sin<span>(<span><span>bx</span><span>−c</span></span>)</span></span></span>+<span>d, where
a is the amplitude
b is the period
c is the phase shift
d is the vertical shift
So, the graph y=3sinx is phase shifted. The phase shift can be calculated as c/b= pi/3/1=pi/3
So, the function is phase shifted for pi/3.</span>

4 0

1 year ago

Three times the sum of half Carlita's age and 3 is at least 12. What values represent Carlita's possible age?

cestrela7 [59]

Answer:

at least 6

Step-by-step explanation:

3(.5x+3)>12

1.5x+9>12

1.5>9

x>6

5 0

1 year ago

Read 2 more answers