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3 years ago

What is the radius of a circumference 56.5

Mathematics

1 answer:

Bas_tet [7]3 years ago

7 0

8.992254284692088 is the radius I believe

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Can you explain what went wrong and the actual answers these 3

vlada-n [284]

For the first one, they got rid of the 6x when they shouldn’t have, and the correct answer is x = 21.5

For the second one, 47 would be subtracted by 180, not 90, and the correct answer is 133

For the third one, 5x - 10 doesn’t equal 115, 5x - 10 equals 180 - 115, and the correct answer is x = 17

7 0

3 years ago

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Asta made a fraction number line to help her multiply 3 x 4/5

aalyn [17]

3 times 4/5 can also be known as 3 times 4 over 5, which is 12 over 5.
12/5 can be simplified to 2 2/5 or 2.4. This is the answer to 3 * 4/5
And I have no idea how to draw a number line on a computer.

7 0

3 years ago

Someone help please! I put A but I’m not sure if it’s correct.

pentagon [3]

Answer: B

21x^2-17x+2

Step-by-step explanation:

4 0

4 years ago

1) A traditional die is a cube with each of its six sides representing the numbers from 1 to 6. Mr. Dicey customized a die by re

liraira [26]

Answer:

1) $\frac{2}{5}$

2) 49.225

3) $\frac{7}{2}$

Step-by-step explanation:

1) To find the expected value of the dice we can use the following equation:

$E(x)=x_{1}*P(x_{1})+x_{2}*P(x_{2})+...+x_{n}*P(x_{n})$

So in our problem the values x will be: 1/1, 1/2, 1/3, 1/4, 1/5 and 1/6 and the probavility for all values is 1/6 so the expected values will be: $E(x)=(\frac{1}{1} *\frac{1}{6}) +(\frac{1}{2} *\frac{1}{6}) +(\frac{1}{3} *\frac{1}{6})+(\frac{1}{4} *\frac{1}{6})+(\frac{1}{5} *\frac{1}{6})+(\frac{1}{6} *\frac{1}{6})$

$E(x)=0.167+0.083+0.056+0.042+0.033+0.028=0.409\approx \frac{2}{5}$

2) To find the variance of the expected values we can use the equation:

$Var(x)=\frac{\sum_{i=1}^{n}(x_{i}-\overline{x})^{2} }{n}$

So for our problem will be:

$Var(x)=\frac{(3-10.5)^2+(4-10.5)^2+(17-10.5)^2+(18-10.5)^2}{4}$

$Var(x)=\frac{56.25+42.25+42.25+56.25}{4}$ $Var(x)=\frac{196.9}{4}=49.225$

3) To find the expected value of the dice we can use the following equation:

$E(x)=x_{1}*P(x_{1})+x_{2}*P(x_{2})+...+x_{n}*P(x_{n})$

So in our problem the values x will be: 1, 2, 3, 4, 5 and 6 and the probavility for all values is 1/6 so the expected values will be: $E(x)=(1*\frac{1}{6}) +(2 *\frac{1}{6}) +(3 *\frac{1}{6})+(4 *\frac{1}{6})+(5 *\frac{1}{6})+(6 *\frac{1}{6})$

$E(x)=0.17+0.33+0.5+0.67+0.83+1=3.5\approx \frac{7}{2}$

7 0

3 years ago

Helppppppppppppppppppp

Karo-lina-s [1.5K]

Answer:

32t+12

Step-by-step explanation:

8 0

3 years ago

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