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1 year ago

Maura swims the length of a pool 8 times .The pool is 17.4 meters long .How many meters does Maura swim in all

Mathematics

1 answer:

Brums [2.3K]1 year ago

3 0

Answer:

139.2

Step-by-step explanation:

8x17.4=139.2

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At the frozen yogurt shop, a machine fills cups with 4 ounces of frozen yogurt before adding the toppings. After the cups are fi

timofeeve [1]

In this question, every cups will be filled with 4 ounces yogurt. That mean, the lowest possible of the cups weight would be 4 ounces. After that the customer can the topping without exceeding 6 ounces of total weight. Since it total weight, that means from the 6 ounces there should be 4 ounces of yogurt. Then the maximum weight is 6 ounces
Minimum weight is 4 ounces and maximum weight is 6 ounces, so the answer would be 4,5,6 or any number between 4-6

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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

1 year ago

Which polynomial correctly combines the like terms and expresses the given polynomial in standard form?

dimaraw [331]

Answer:

D) n^6 - 6m^6 + 7mn^5 + 14m^2n^4 -5m^3 n^3

Step-by-step explanation:

The given polynomial is 8mn^5 -2m^6 +5m^2 n^4 - m^3 n^3 + n^6 - 4m^6 + 9m^2n^4 -mn^5 - 4m^3n^3

Now we have to identify the like terms and simplify.

Like terms are nothing but the terms which have the same variables with same powers.

= (8mn^5 - mn^5) -2m^6 - 4m^6 + 5m^2 n^4 + 9m^2 n^4 - m^3n^3 - 4m^3n^3 +n^6

Now combine the like terms

= 7mn^5 - 6m^6 +14m^2 n^4-5m^3n^3 + n^6

It can be written in standard form

= n^6 - 6m^6 + 7mn^5 + 14m^2n^4 -5m^3 n^3

The answer is D)

Hope this will helpful.

Thank you.

6 0

1 year ago

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What is the degree of the sin of .19

Tpy6a [65]

Sin(19)

Use your calculator.

sin(19) = 0.3255681545

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1 year ago

The selling price of a box of crackers is $1.75 You mark the crackers up to $2.54 . What is the markup percentage?

Aleks04 [339]

The markup percentage is 45.14%

Step-by-step explanation:

The given is:

The selling price of a box of crackers is $1.75
You mark the crackers up to $2.54

We need to find the markup percentage

The markup percentage = $\frac{New-Old}{old}$ × 100%

∵ The selling price of a box of crackers is $1.75

∴ Old = 1.75

∵ You mark the crackers up to $2.54

∴ New = 2.54

- Substitute these values in the rule above

∵ The markup percentage = $\frac{2.54-1.75}{1.75}$ × 100%

∴ The markup percentage = $\frac{0.79}{1.75}$ × 100%

∴ The markup percentage = 0.4514 × 100%

∴ The markup percentage = 45.14%

The markup percentage is 45.14%

Learn more:

You can learn more about percentage in brainly.com/question/1834017

#LearnwithBrainly

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2 years ago

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