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8_murik_8 [283]
1 year ago
5

A basketball court is a rectangle that is 28\,\text{m}28m28, start text, m, end text long and 15\,\text{m}15m15, start text, m,

end text wide. A diagram on a coordinate plane shows a circle in the middle of the court centered at (0\,\text{m},0\,\text{m})(0m,0m)left parenthesis, 0, start text, m, end text, comma, 0, start text, m, end text, right parenthesis. The point (0.8\,\text{m}, 1.5\,\text{m})(0.8m,1.5m)left parenthesis, 0, point, 8, start text, m, end text, comma, 1, point, 5, start text, m, end text, right parenthesis is on the circumference of the circle.
Approximately what percentage of the court does the circle cover?
Mathematics
1 answer:
vladimir2022 [97]1 year ago
7 0

Answer:

48

Step-by-step explanation:

You might be interested in
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&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.1(1.62)+(1-0.1)(1.80) \\  \\ &#10;=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&#10; A_{10}+(1-\alpha)F_{10} \\  \\ =0.3(1.57)+(1-0.3)(1.76) \\  \\ &#10;=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&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.3(1.62)+(1-0.3)(1.70) \\  \\ &#10;=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&#10; A_{10}+(1-\alpha)F_{10} \\  \\ =0.5(1.57)+(1-0.5)(1.72) \\  \\ &#10;=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&#10; A_{11}+(1-\alpha)F_{11} \\  \\ =0.5(1.62)+(1-0.5)(1.65) \\  \\ &#10;=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:

&#10; \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = &#10;\frac{|-0.17|+|-0.08|+|-0.07|}{3}  \\  \\ = \frac{0.17+0.08+0.07}{3} = &#10;\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:

&#10; \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = &#10;\frac{|-0.15|+|-0.03|+|0.11|}{3}  \\  \\ = \frac{0.15+0.03+0.11}{3} = &#10;\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
Aiden borrows a book from a public library. He read a few pages on day one. On day two, he reads twice the number of pages than
Fofino [41]

Let the number of pages read on day 1 be = x

Then on day 2 he read twice the pages from day one, it is = 2x

On 3rd day he read 6 pages less than 1st day, it is = x-6

Total pages are = 458

The equation becomes: x+2x+x-6=458

4x-6=458

4x=464

x=116

So on the third day, Aiden read x-6=116-6=110 pages.

5 0
1 year ago
a rectangular piece of paper 71 cm long and 10m wide . a cylinder is formed by rolling the paper along its length , find volume
natka813 [3]

Answer:

Volume=355 cm³

Step-by-step explanation:

I assume it should be 10 cm, not 10 m.  If it is 10 m, everything will have to be adjusted.

Circumference = 2\pir

10=2\pir

10÷2\pi=r

1.59=r

V=\pir²h

V=\pi(1.59)(71)

v=355

7 0
1 year ago
Celia bought a bag of 121212 goldfish for \$3$3dollar sign, 3.
muminat

Celia bought a bag of 12 goldfish for $3,  Cost of 1 goldfish is $0.25 .

<u>Step-by-step explanation:</u>

The goldfish is a freshwater fish in the family Cyprinidae of order Cypriniformes. It is one of the most commonly kept aquarium fish. A relatively small member of the carp family, the goldfish is native to East Asia.

Here we have , Celia bought a bag of 12 goldfish for $3 . We need to find the cost of 1 goldfish . Let's find out:

Celia have a bag of 12 goldfish which cost her $3 . So , cost of 1 goldfish is equivalent to

⇒ (1)\frac{12}{3}

⇒ (1)\frac{3(4)}{3}

⇒ \frac{1}{4}

⇒ 0.25

Therefore, Celia bought a bag of 12 goldfish for $3  , Cost of 1 goldfish is $0.25 .

5 0
1 year ago
Two boats leave port at noon. Boat 1 sails due east at 12 knots. Boat 2 sails due south at 8 knots. At 2 pm the wind diminishes
Ivan

Answer:

14.86 knots.

Step-by-step explanation:

<em>Given that:</em>

The boats leave the port at noon.

Speed of boat 1 = 12 knots due east

Speed of boat 2 = 8 knots due south

At 2 pm:

Distance traveled by boat 1 = 24 units due east

Distance traveled by boat 2 = 16 units due south

Now, speed of boat 1 changes to 9 knots:

At 3 pm:

Distance traveled by boat 1 = 24 + 9= 33 units due east

Distance traveled by boat 2 = 16+8 = 24 units due south

Now, speed of boat 1 changes to 8+7 = 15 knots

At 5 pm:

Distance traveled by boat 1 = 33 + 2\times 9= 51 units due east

Distance traveled by boat 2 = 24 + 2 \times 15 = 54 units due south

Now, the situation of distance traveled can be seen by the attached right angled \triangle AOB.

O is the port and A is the location of boat 1

B is the location of boat 2.

Using pythagorean theorem:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}\\\Rightarrow AB^{2} = OA^{2} + OB^{2}\\\Rightarrow AB^{2} = 51^{2} + 54^{2}\\\Rightarrow AB^{2} = 2601+ 2916 = 5517\\\Rightarrow AB = 74.28\ units

so, the total distance between the two boats is 74.28 units.

Change in distance per hour = \dfrac{Total\ distance}{Total\ time}

\Rightarrow \dfrac{74.28}{5} = 14.86\ knots

6 0
1 year ago
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