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3 years ago

What’s the length of the wall?

Mathematics

2 answers:

irinina [24]3 years ago

5 0

Answer:

Step-by-step explanation:

Allushta [10]3 years ago

5 0

A is the answer
have a good day!

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Given: m<4 + m<7 = 180° Prove: c ll d I need the statements and reasons

AVprozaik [17]

The horses name was friday

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2 years ago

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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

3 years ago

How is number 0.000063 written in scientific notation?

katen-ka-za [31]

Step-by-step explanation:

The scientific notation:

$a\times10^k$

1 ≤ a < 10 and k - any integer number

$0.000063=0\underbrace{.00006}_{5\to}3=6.3\times10^{-5}\\\\\\0.000063=6.3:100,000=6.3:10^5=6.3\times10^{-5}$

5 0

3 years ago

In 2003 there were 1078 JC Penney stores and in 2007 there were 1067 stores.

Tamiku [17]

Answer:

a) $f(s)=\frac{-11}{3}s+\frac{3245}{3}$ $f(1)=1078 stores$ b) 2012, 1049 stores. 2014, 1041 stores. c) Yes it is a downsizing company.

Step-by-step explanation:

s=year | f(s) = number of stores per year

1 | 1078

4| 1067

$m=\frac{1067-1078}{4-1}\Rightarrow m=\frac{-11}{3}\\1078=-\frac{11}{3}*1+b\Rightarrow b=\frac{3245}{3}\\f(s)=\frac{-11}{3}s+\frac{3245}{3}$

s=1 in 2003

$s=1 year =2003\\ Testing\\f(1)=\frac{-11}{3}(1)+\frac{3245}{3}=\frac{3234}{3}=1078\\ f(1)=1078$

b) For 2012, s=9. For 2014, s=11

$\\f(s)=\frac{-11}{3}s+\frac{3245}{3}\\ \\f(9)=\frac{-11}{3}(9)+\frac{3245}{3}\cong1049\\\\f(11)=\frac{-11}{3}(11)+\frac{3245}{3}\cong=1041$

c) In deed. Since the function shows a decreasing amount of JC Penney stores. Maybe this is a downsizing from this company, progressively closing some stores.

4 0

4 years ago

Casper has some whipping cream that is 18%, percent butterfat, and some milk that is %4, percent butterfat. He wants to make a 5

steposvetlana [31]

L represent Option A. The correct answer is that the mixture has the intended volume and has more than the intended percent of butterfat.

<h3>What is the graph about?</h3>

Note that:

The graph shows that the point L is one that is below the blue line of the graph.

Therefore, x + y = 500

Also one need to see that:

0.18x + 0.04y = 500(0.12)

Since that cream is 18% butterfat, milk is 4% butterfat, so L is composed of of 350 ml of cream and 100 ml of milk.

Hence the total amount of butterfat for L is:

0.18 x 350 + 0.04 x 100

= 67ml.

Next, one has to calculate 12% of the volume of L :

12% of 450:

0.12 x 450 = 54ml

So, because 67 is bigger than 54, this implies that the volume of butter fat in L is more than 12 %.