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forsale [732]
3 years ago
10

What is the area of a circle with diameter 8cm? Round your answer to the nearest tenth. use pi= 3.14

Mathematics
2 answers:
hodyreva [135]3 years ago
7 0

Answer:

50.2 square cm

Step-by-step explanation:

Diameter = 8 cm

Radius = 8/2 = 4 cm

area \: of \: circle = \pi {r}^{2}  \\  = 3.14 \times  {4}^{2}  \\  = 3.14 \times 16 \\  = 50.24 \:  {cm}^{2}  \\  = 50.2 \:  {cm}^{2}

balandron [24]3 years ago
7 0

Answer:

50.62=51

Step-by-step explanation:

A=pi*r^2

=3.14*(4^2)

=50.62=51

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<span>Pearson's correlation coefficient 'r' measures the degree to which two variables are correlated. The value is between -1 and 1. Negative values correspond to negative correlation (as one variable increases, the other variable decreases) whereas positive values correspond to positive correlation. The larger the absolute value of r, the closer a correlation graph resembles a set of points on the same line. Therefore, if two variables are completely correlated, the correlation value is 1 or -1, depending on whether the correlation is positive or negative.</span>
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A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by th
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Step-by-step explanation:

I assume that "ground" is at 0 ft height. which is in an actual scenario not airways the case.

y = -16x² + 64x + 89

shows us that the tower is 89 ft tall (the result for x = 0, at the start).

anyway, if the original assumption is correct, then we need to solve

0 = -16x² + 64x + 89

the general solution for such a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/)2a)

in our case

a = -16

b = 64

c = 89

x = (-64 ± sqrt(64² - 4×-16×89))/(2×-16) =

= (-64 ± sqrt(4096 + 5696))/-32 =

= (-64 ± sqrt(9792))/-32

x1 = (-64 + 98.95453501...)/-32 = -1.092329219... s

x2 = (-64 - 98.95453501...)/-32 = 5.092329219... s

the negative solution for time is but useful here (it would be the time calculated back to ground at the start).

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2 years ago
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Answer:

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Step-by-step explanation:

8 0
2 years ago
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teacher gives 5 students a multiple choice test, in which each problem is worth 1 point and there is no penalty with negative po
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The mean is the average value of a given set of numbers, and the median is the number in the middle of a set of numbers arranged in increasing order

The correct values as response to the questions are;

The minimum possible <em>top score</em> is <u>21</u>

The maximum possible <em>top score</em> is <u>32</u>

The <em>minimum </em>of the possible <em>standard deviation</em> is approximately <u>1.26</u>

The <em>maximum </em>of the possible <em>standard deviation</em> is approximately <u>11.7</u>

<u />

The reason the above values are correct are as follows:

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The given parameters are;

The number of students that take the test, n = 5

The amount of points for each problem = 1 point

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The mean Score = 10

Required:

The minimum possible score;

The maximum possible top score

The minimum of the possible standard deviations

The maximum of the possible standard deviations

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Given that there is a score (the median) which is 9, we have;

The scores obtainable in the test is Scores ≥ 9

Therefore, for a score of 10, we have;

The minimum total points obtainable = Mean × Number of students

Therefore;

Total minimum total points obtainable by the 5 students = 5 × 10 = 50

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By arrangement, with the median at the middle, the minimum possible top score is given as follows;

0, 0, 9, 20, 21

Therefore, the minimum possible top score is <u>21</u>

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  • The maximum possible top score

The maximum possible top score is similarly given by arrangement of the numbers as follows'

0, 0, 9, 9, 32

The maximum possible top score is <u>32</u>

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  • The minimum standard deviation

By arrangement and selection, the minimum standard deviation is given as follows;

The minimum of the possible standard deviations of (9, 9, 9, 11, 12) ≈ <u>1.26</u>

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The maximum of the possible standard deviation is given as follows;

The maximum of the possible standard deviation of (0, 0, 9, 9, 32) ≈ <u>11.7</u>

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Learn more about mean and median here:

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