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2 years ago

(iii) Find the smallest whole number h such that4704/his a cube number.

Mathematics

1 answer:

yan [13]2 years ago

6 0

Answer:

$h=588$

Step-by-step explanation:

Given the fraction $\dfrac{4,704}{h}$

Consider its numerator and factor it:

$4,704=2^5\cdot 3\cdot 7^2$

You cann not take the cube root from $3, 7^2, 2^5,$ you can only take the cube root from $2^3$ , then the smallest whole number $h$ such that $\dfrac{4,704}{h}$ is a cube number is

$h=2^2\cdot 3\cdot 7^2=4\cdot 3\cdot 49=588$

When $h=588,$

$\dfrac{4,704}{588}=8=2^3$

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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

NEED HELP FAST WILL GIVE BRAINIEST

vladimir1956 [14]

Answer:

x^2 + 8x - 65 = 0.

Step-by-step explanation:

In order to solve the side of the enlarged area of 81 square inches, we use the equation in solving the area of a square. Area = side^2.

the increase in side will be x, so (4+x)^2 = 81, x^2 + 4x + 4x + 16 = 81

x^2 + 8x + 16 - 81 = 0

x^2 + 8x - 65 = 0.

Hoped this helped mark Brainliest!

5 0

1 year ago

Read 2 more answers

Jayden has some dimes and some quarters. He has at most 25 coins worth at least $4.60 combined. If Jayden has 7 dimes, determine

Y_Kistochka [10]

Answer:

Step-by-step explanation:

A dime is worth 10 cents. Converting to dollars, it becomes 10/100 = $0.1

A quarter is worth 25 cents. Converting to dollars, it becomes 25/100 = $0.25

Let x represent the number of dimes that Jayden has.

Let y represent the number of quarters that Jayden has.

Jayden has some dimes and some quarters. He has at most 25 coins. It means that

x + y ≤ 25

The coins worth at least $4.60 combined. It means that

0.1x + 0.25y ≥ 4.6 - - - - - - - - - - 1

If Jayden has 7 dimes, then

7 + y ≤ 25

y ≤ 25 - 7

y ≤ 18

Substituting x = 7 into equation 1, it becomes

0.1 × 7 + 0.25y ≥ 4.6

0.7 + 0.25y ≥ 4.6

0.25y ≥ 4.6 - 0.7

0.25y ≥ 3.9

y ≥ 3.9/0.25

y ≥ 15.6

All possible values for the number of quarters that he could have would be

15.6 ≤ y ≤ 18

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2 years ago

A drawing of a surfboard in a catalog shows its length as 8 4/9 inches. Find the actual length of the surfboard if 1/2 inch leng

jonny [76]

3/8 of a foot = 3/8 x 12/1 or 4 1/2 inches

So our 1/2 inch grows by a factor of 9 because 4 1/2 ÷ 1/2 = 9 Think how many 1/2 dollars are in 4 1/2 dollars. (Answer is 9)

So our 8 4/9 in the catalog has to grow the same

8 4/9 x 9 = 8 4/9 x 9/1 = 76/9 x 9/1 or 76 inches which is 6 ft 4 inches.
(76 ÷12 = 6 r4)

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2 years ago

Need help RN!!!!

Vesnalui [34]

Answer:

log y = 3

Step-by-step explanation:

Substitute 10 for x, obtaining the following:

log y = 0.23(10) + 0.8, or

log y = 2.3 + 0.8, or 3.10, or (to the nearest whole number):

log y = 3

Step-by-step explanation:

Hope this helps:)

7 0

1 year ago

(iii) Find the smallest whole number h such that4704/his a cube number.​

(iii) Find the smallest whole number h such that4704/his a cube number.