All categories

3 years ago

Write the fraction 8/9 as a percent round to the nearest hundredth of a percent were necessary?

Mathematics

1 answer:

Flura [38]3 years ago

7 0

Answer:

Step-by-step explanation:

To turn a fraction into a percent, you divide the numerator by the denominator.

8 divided by 9 = .8 repeated

To turn it into a percent, you move the decimal to the right 2 spaces.

.8 turns into 88.8 repeated

Then you round to the nearest hundredth

88.8 repeated rounds to 88.89.

You might be interested in

Consider a circle centered at C(-2,-4). If the point P(1,-1) lies on the circle, then which of the following points also lies on

Aliun [14]

Answer:

Consider a circle centered at C(-2,-4). If the point P(1,-1) lies on the circle, then which of the following points also lies on the circle?

B. (-2+√18, -4)

C. (√18, -4)

D. (-2, 4+√18)

Step-by-step explanation:

If the distance is greater than the radius, the point lies outside. If it's equal to the radius, the point lies on the circle. And if it's less than the radius, you guessed it right, the point will lie inside the circle.

hope it's help you

7 0

3 years ago

The volume of a cone varies jointly with the area of the base and the height. When the area of the base is 27 cm2 and the height

Tomtit [17]

Cone=cup:
V=124cm^3.
h=12cm
V=Bh
P=r^2π+rsπ:
B=r^2π
B=V/h=124/12=10,3cm^2
r^2=B/π
r=√(B/π)=1,81 cm
S=√(r^2+h^2)=√(12^2+1.81^2)=√(144+3,28)=12,13 cm
P=r^2π+rSπ=3,28*3,14+1,81*12,13*3,14=10,29+68,93=79,22 cm^2

3 0

3 years ago

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 inequality represents this sentence? Seven plus five is greater than six plus two. A. 7 + 5 ≥ 6 + 2 B. 6 + 2 > 7 + 5 C.

Ainat [17]

The answer would be D

Hope this helps!

5 0

3 years ago

Read 2 more answers

The decimal d is formed by writing in succession all the positive integers in increasing order after the decimal point; that is,

Maksim231197 [3]

Answer:

Step-by-step explanation:

We are given that a decimal number d

d=0.12345678...

Where d is formed by writing in succession all the positive integers in increasing order after decimal point.

We have to find the 100th digit of d to the right of the decimal point.

Place of first digit 1 after decimal=Tenth

Place of second digit 2 after decimal=Hundredth

Place of third digit 3 after decimal=Thousandth

Therefore, 100th digit of d to the right of the decimal point=2

5 0

3 years ago