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2 years ago

A staircase has 105 blocks, how many stairs does it have?

1 answer:

7 0

1 1
2 3
3 6
4 10
5 15
6 21
7 28
8 36
9 45
10 55
11 66
12 78
13 91
14 105
15 120
16 136
17 153
18 171
19 190
20 210
21 231
22 253
23 276
24 300

here is a spreadsheet to help you understand the stair problem notice for 1 step 1 block, 2 step, 3 block, 3 step, 6 block and so on notice the blocks for the number of a step is that step plus all the other blocks needed to make all the one(s) below it. Notice it takes 105 blocks to make 14 steps.
So if you are familiar with the summation process you would add the number that go with that stair with all the stairs below it. example step 5 blocks would be 5 + 4 + 3 + 2 +1 or 15 blocks. I hope this gets you started. The second part I do not understand the problems because they are not equations. NO equal signs

glad if it helps

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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

Which transformation best describes the image of a object viewed through a microscope?

algol13

A. dilation.............

4 0

2 years ago

Read 2 more answers

Becky adds 3/4 gallon of gasoline into her snow thrower's empty fuel tank. She uses 1/4 gallon for the snowstorm on Friday, and

Kaylis [27]

Amount of fuel in the tank before the story begins . . . . . zero.

Total amount poured in . . . . . 3/4 gallon.

Total amount burned:
   on Friday . . . . . 1/4 gallon
   on Sunday . . . . 1/4 gallon
   Total used . . 1/2 gallon .

Amount remaining in the tank on Monday:

   (3/4 gallon in) - (1/2 gallon burned) = 1/4 gallon left.
      ==> NOT empty

The tank would have been empty on Monday IF Becky
had poured in only 1/2 gallon, instead of 3/4 of a gallon
before the first flakes began to fly.

8 0

2 years ago

Maria records random speeds from three different Internet providers in the table. Provider Download Speed (megabits per second)

Igoryamba

Answer:

Able Cable

Step-by-step explanation:

Mean Value of Samples

It's central tendency statistics measure that gives one representative value out of a complete dataset. It can be computed as the sum of the values by the number of values:

$\displaystyle \bar x=\frac{\sum x_i}{n}$

We need to evaluate the mean value of the three internet providers mentioned in the question.

For CityNet:

$\displaystyle \bar x_1=\frac{3.6+3.7+3.7+3.6+3.9}{5}=3.7$

For Able Cable:

$\displaystyle \bar x_2=\frac{3.9+3.9+4.1+4.0+4.1}{5}=4.0$

And for Tel-N-Net:

$\displaystyle \bar x_3=\frac{3.9+3.7+4.0+3.6+3.8}{5}=3.8$

We can conclude Able Cable offers the fastest mean downloading internet speed

7 0

2 years ago

Given H(6,7) and I(-7,-6), if point G lies 1/2 of the way along line segment HI, Santiago argues that point G is located at the

irinina [24]

Given the points H(6,7) and I(-7,-6).

If point G lies $\frac{1}{2}$ of the way along line segment HI.

Therefore, we can say that the point G divides the line segment HI in the ratio 1:1.

So, by using the cross section formula we can determine the coordinates of point G.

For the given points say $(x_1, y_1)$ and $(x_2, y_2)$ divided is in the ratio $m_1 : m_2$ , the coordinates are $(\frac{m_1x_2+m_2x_1}{m_1+m_2} , \frac{m_1y_2+m_2y_1}{m_1+m_2} )$

Coordinates G = $(\frac{(1 \times -7)+(1 \times 6)}{2} , \frac{(1 \times -6)+(1 \times 7)}{2})$

= (-0.5 , 0.5)

Hence, the coordinates of G are (-0.5 , 0.5).

So, Santiago argues that point G is located at the origin. The point G is located at (-0.5, 0.5). Therefore, he is not correct.

8 0

2 years ago