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Dmitriy789 [7]
3 years ago
6

A salesperson at a jewelry store earns ​5% commission each week. Last​ week, Jarrod sold $640 worth of jewelry. How much did he

make in​ commission?
Mathematics
2 answers:
Masteriza [31]3 years ago
4 0

Answer:

It $32

$540x.05=$27

21.60!

Step-by-step explanation:

            it just a tip

Heidi made $32 from The commission, And the store made $513 (i had this question myself on Savvas)

it just a tip

1. 540x0.05=503

2. 540-37= $513

So the store made 513 dollars and Heidi made 27 dollars

5% = 5/100 = 0.05 = 1/20

Then 5% × $640 = $640/20 = $64/2 = $32.

5% × $640 = $32

Aleksandr-060686 [28]3 years ago
3 0

Answer:

  $32

Step-by-step explanation:

To find the commission, multiply the rate by the sales:

  commission = 5% × $640 = $32

He made $32 in commission.

__

<em>Additional comment</em>

Many calculators and all spreadsheets can work with percentages directly. If yours can't, or if you can't do this arithmetic mentally, you can use the equivalent decimal or fraction for 5%.

  5% = 5/100 = 0.05 = 1/20

Then 5% × $640 = $640/20 = $64/2 = $32.

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Hiros family lives 448 kilometers from the beach.Each of the 5 adults drove the family van an equal distance to get to and from
Drupady [299]

Answer:

179.2km

Step-by-step explanation:

The distance from their house to the beach is 448km. Now they have to drive to and from the beach. The total distance traveled by the family is 448km + 448km.

This is equal to 896km. Now we have 5 adults who took their turns to drive and they drove the same distance. The total distance traveled by each adult will be 896/5 = 179.2km

Hence, each adult in the family drove a distance of 179.2km

3 0
3 years ago
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}&#10;{|c|c|c|c|}&#10;Month&Price per Chip&Month&Price per Chip\\[1ex]&#10;January&\$1.90&July&\$1.80\\&#10;February&\$1.61&August&\$1.83\\&#10;March&\$1.60&September&\$1.60\\&#10;April&\$1.85&October&\$1.57\\&#10;May&\$1.90&November&\$1.62\\&#10;June&\$1.95&December&\$1.75&#10;\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&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.1(1.62)+(1-0.1)(1.80) \\  \\ &#10;=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&#10; A_{10}+(1-\alpha)F_{10} \\  \\ =0.3(1.57)+(1-0.3)(1.76) \\  \\ &#10;=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&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.3(1.62)+(1-0.3)(1.70) \\  \\ &#10;=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&#10; A_{10}+(1-\alpha)F_{10} \\  \\ =0.5(1.57)+(1-0.5)(1.72) \\  \\ &#10;=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&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.5(1.62)+(1-0.5)(1.65) \\  \\ &#10;=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:

&#10; \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = &#10;\frac{|-0.17|+|-0.08|+|-0.07|}{3}  \\  \\ = \frac{0.17+0.08+0.07}{3} = &#10;\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:

&#10; \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = &#10;\frac{|-0.15|+|-0.03|+|0.11|}{3}  \\  \\ = \frac{0.15+0.03+0.11}{3} = &#10;\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
3 years ago
Is the figure an example of a segment?
Radda [10]

Answer: yes, it extends infinitely in opposite directions

Step-by-step explanation:

Examples of line segments include the sides of a triangle or square. More generally, when both of the segment's end points are vertices of a polygon or polyhedron, the line segment is either an edge (of that polygon or polyhedron) if they are adjacent vertices, or otherwise a diagonal.

Let me know if this helped

5 0
3 years ago
Find the area and the circumference of a circle with the radius 3 ft.Use 3.14 for pi
Lilit [14]

ANSWER

• Circumference: 18.84 ft

,

• Area: 28.26 ft²

EXPLANATION

The circumference of a circle of radius r is,

C=2\pi r

In this case, the radius is r = 3 ft and we have to use 3.14 for π,

C=2\cdot3.14\cdot3ft=18.84ft

Hence, the circumference of this circle is 18.84 feet.

The area of a circle of radius r is,

A=\pi r^2

In this case, r = 3 ft and π = 3.14,

A=3.14\cdot3^2ft^2=28.26ft^2

Hence, the area of this circle is 28.26 square feet.

6 0
1 year ago
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