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2 years ago

Calculate a23 for the product of the following matrices

Mathematics

1 answer:

s344n2d4d5 [400]2 years ago

6 0

If $A$ is the matrix product

$A=\underbrace{\begin{pmatrix}1&2\\4&3\end{pmatrix}}_{M_1}\underbrace{\begin{pmatrix}6&1&5\\7&3&4\end{pmatrix}}_{M_2}$

and $a_{2,3}$ is the entry of $A$ in the second row and third column, then

$a_{2,3}=\begin{pmatrix}4&3\end{pmatrix}\begin{pmatrix}5\\4\end{pmatrix}=4\times5+3\times4=\boxed{32}$

That is, the 2nd row, 3rd column entry of $A$ is the product of the 2nd row of $M_1$ and the 3rd column of $M_2$ .

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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 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:

Given 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

Part 2A:

Given 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


Part 2B:

Given 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


Part 3A:

Given 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


Part 3B:

Given 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 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 using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157

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 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 using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107


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 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 using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097

5 0

2 years ago

What is the maximum number of pages the report can have so that it can be completely stored on a USB drive that holds 4 \cdot 10

lana [24]

Answer:

2,300,000pages report.

Step-by-step explanation:

Given a USB drive with memory of 4.6*10⁶KB, in order to know the maximum number of pages a report can have so that it can be completely stored on the drive, the conversion factor must be used.

We must first understand that 1 page of a document uses up approximately 2kB of the storage.

Since 2kB = 1page

4.6*10⁶KB = x page

Cross multiply

2kB * x = 4.6*10⁶KB * 1

2x = 4.6*10⁶

x = 4.6*10⁶/2

x = 2.3*10⁶

x = 2,300,000

Hence the maximum number of pages that the report can have so that it can be completely stored on a USB drive that holds 4.6*10⁶KB is approximately 2,300,000pages report.

4 0

2 years ago

Which graph represents a function?

Agata [3.3K]

The fourth one because to know if it is a function or not it has to pass the vertical line test the first one when you draw a vertical line some of the lines repeats so does the second and third but the last one when you draw a vertical line it doesn't repeat s that means its a function

8 0

2 years ago

Read 2 more answers

Desiree works 28 hours per week. She has a monthly income of $120 from investments. Desiree also plays in a band one night a wee

Oksi-84 [34.3K]

So hmmm let's see
she has a monthly income of 120 from investments, now, there are 12 months in a year, so her yearly income from investments are 120*12 or
$1440

she plays on a band, and makes 200 a week, now, there are 52 weeks in a year, so her yearly income from band playing is 200 * 52, or
$10400

her total annual income is 49696, now, if we subtract the band and investment income, we'd be left over with only what comes from her job payrate
so 49696 - 1440 - 10400 is 37856

so, she makes from her job, $37856 annually

now, she only works 28 hours weekly, how much is that yearly? well, 52 weeks in a year, she works 28*52 hours a year, let us divide 37856 by that

37856 ÷ ( 28 * 52) well, it ends up as 26

so, her hourly payrate is $26 per hour

now, she wants to ask for a raise, to make 51880 annually

well, if we check the difference of 51880 and 49696, that'd leave us with the difference in pay, or the raise annual amount

51880 - 49696 = 2184

ok, so she wants $2184 annually more from her work
how much is that in the hours she works annually? well 2184 ÷ ( 28 * 52)

7 0

2 years ago

Read 2 more answers

The sanitation department calculated that last year each city resident produced approximately 1.643 × 103 pounds of garbage. The

bija089 [108]

Given that, according to sanitation departments report; the samount of garbage produced by 1 resident $= 1.643 \times 10^3$ pound.