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ale4655 [162]
3 years ago
10

HELP plssssssssssssssssssssssssssssssssssssssssssssssss

Mathematics
2 answers:
V125BC [204]3 years ago
5 0

Answer:18×6=108

D SHOULD BE RIGHT

d is = to 108.

tell me if right

Kay [80]3 years ago
4 0
It is B. Hope this helps
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Help pls ill give Brainlyest to the one who does it step by step.
Elanso [62]

Answer:

12.8% decrease

Step-by-step explanation:

if you find the unit rate for the sale, which would be 125 cents divided by 3 cans, equals around 43.6 cents per can. 43.6 cents is an x% decrease of 50 cents, and to find x easily i multiplied by 2. 43.6 x 2=87.2, and 50 x 2 = 100. that means that the sale price is 87.2 % of the original price, which is a 12.8% decrease.

7 0
2 years ago
Find the product of all positive divisors of 288.
irinina [24]

Answer:

<em>The</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>1</em><em>.</em><em>5</em><em>1</em><em>4</em><em>5</em><em>7</em><em>1</em><em>8</em><em>9</em><em>4</em><em>×</em><em>1</em><em>0</em><em>^</em><em>2</em><em>1</em><em>.</em>

Step-by-step explanation:

<em>Here</em><em>, </em><em> </em><em>the</em><em> </em><em>divisors</em><em> </em><em>of</em><em> </em><em>2</em><em>8</em><em>8</em><em> </em><em>are</em><em>,</em>

<em>1</em><em>,</em><em>2</em><em>,</em><em>3</em><em>,</em><em>4</em><em>,</em><em>6</em><em>,</em><em>8</em><em>,</em><em>9</em><em>,</em><em>1</em><em>2</em><em>,</em><em>1</em><em>6</em><em>,</em><em>1</em><em>8</em><em>,</em><em>2</em><em>4</em><em>,</em><em>3</em><em>2</em><em>,</em><em>3</em><em>6</em><em>,</em><em>4</em><em>8</em><em>,</em><em>7</em><em>2</em><em>,</em><em>9</em><em>6</em><em>,</em><em>1</em><em>4</em><em>4</em><em>,</em><em>2</em><em>8</em><em>8</em><em>.</em>

<em>now</em><em>,</em><em> </em><em>their</em><em> </em><em>product</em><em> </em><em>=</em><em>1</em><em>×</em><em>2</em><em>×</em><em>3</em><em>×</em><em>4</em><em>×</em><em>6</em><em>×</em><em>8</em><em>×</em><em>9</em><em>×</em><em>1</em><em>2</em><em>×</em><em>1</em><em>6</em><em>×</em><em>1</em><em>8</em><em>×</em><em>2</em><em>4</em><em>×</em><em>3</em><em>2</em><em>×</em><em>3</em><em>6</em><em>×</em><em>4</em><em>8</em><em>×</em><em>7</em><em>2</em><em>×</em><em>9</em><em>6</em><em>×</em><em>1</em><em>4</em><em>4</em><em>×</em><em>2</em><em>8</em><em>8</em>

<em>=</em><em>1</em><em>.</em><em>5</em><em>1</em><em>4</em><em>5</em><em>7</em><em>1</em><em>8</em><em>9</em><em>4</em><em>×</em><em>1</em><em>0</em><em>^</em><em>2</em><em>1</em><em>.</em>

<em>is</em><em> </em><em>answer</em><em>. </em>

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

6 0
2 years ago
Read 2 more answers
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Juli2301 [7.4K]
Its true because you have to subtract in the long version so do you want me to show it to you?

3 0
3 years ago
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What is the area of these trapezoid
Assoli18 [71]

Answer:

1. 54ft squared  2. 48 cm squared

Step-by-step explanation:

6 0
2 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
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