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2 years ago

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A group of astronomers observed light coming from a star located a distance of 331,000,000,000,000,000,000,000,000 light-years f

rom Earth. What is this distance expressed in scientific notation?

2 answers:

Nikitich [7]2 years ago

8 0

3.31 x 10^26 im pretty sure

Anika [276]2 years ago

5 0

3.31 times 10 to the 26th power

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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 <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:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Therefore, the foreast for period 12 is $1.78</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:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70

</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:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68

</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:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65

</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:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

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 </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:

$
 \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 </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:

$
 \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 </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>

5 0

1 year ago

If Aiden and Natalie each set the arm length of their catapults to 55 centimeters, which statement is true?

Tema [17]

Answer:

Not enough information

Step-by-step explanation:

one of the catapults could have more force than the other

7 0

1 year ago

Read 2 more answers

On a dell river, a boat will pass the Colby drawbridge and then the wave drawbridge opening times.There are 60 minutes in a hour

Ksju [112]

45/60 and 10/60 because we know that the common denominator is 60 then 3/4 of an hour is 45mins then 1/6 of an hour is 10

6 0

1 year ago

Read 2 more answers

Suppose 50 percent of the customers at Pizza Palooza order a square pizza, 70 percent order a soft drink, and 35 percent order b

Firdavs [7]

Answer:

Ordering a soft drink is independent of ordering a square pizza.

Step-by-step explanation:

20% more customers order a soft drink than pizza, therefore they cannot be intertwined.

Given: P(A)=0.5 & P(B)=.7

P(A∩B) = P(A) × P(B)

= 0.5 × .7

= 0.35

P(A∪B) = P(A) + P(B) - P(A∩B)

= 0.5 + .7 - 0.35

= 0.85

P(AΔB) = P(A) + P(B) - 2P(A∩B)

= 0.5 + .7 - 2×0.35

= 0.5

P(A') = 1 - P(A)

= 1 - 0.5

= 0.5

P(B') = 1 - P(B)

= 1 - .7

= 0.3

P((A∪B)') = 1 - P(A∪B)

= 1 - 0.85

= 0.15

7 0

1 year ago

Sierra is selling bracelets to raise money for the Language Club at school. Each bracelet with yellow beads sells for $5. Each b

matrenka [14]

Answer:

Correct answer is:

$5y+6r=660\\y=2r+8$

Step-by-step explanation:

Given that Number of bracelets with yellow beads is represented by $y$

Each bracelet with yellow beads is sold for $5.

Total money raised by bracelets with yellow beads = Number of bracelets sold $\times$ Money raised by sale of one such bracelet = $5y$

Also Given that Number of bracelets with Orange beads is represented by $r$

Each bracelet with orange beads is sold for $6.

Total money raised by bracelets with orange beads = Number of bracelets sold $\times$ Money raised by sale of one such bracelet = $6r$

Given that total money raised by sale of both type of bracelets is $660.

so, the first equation becomes:

$5y+6r=660 ....... (1)$

It is also given that "<em>The number of bracelets with yellow beads that Sierra sold is 8 more than twice the number of bracelets with orange beads</em>"

$\Rightarrow r =2r+8 ...... (2)$

So, by equation (1) and (2), the system of equations is:

$5y+6r=660\\y=2r+8$

5 0

2 years ago