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tiny-mole [99]
2 years ago
5

Kasha has a spinner divided into 8 equal sections that are labeled 1 though 8. She wants to compare the theoretical probability

and
the experimental probability of spinning an odd number. She spins the spinner 10 times and records the results in this list.
{2, 4, 1, 8, 7,5, 3, 4, 1,5}
Drag and drop the answers into the boxes to correctly complete the sentences
The theoretical probability of spinning an odd number is equal to
The experimental probability of spinning an odd
number is equal to
Therefore, the theoretical probability of spinning an odd number is
the
experimental probability of spinning an odd number
3
5
greater than
less than
Mathematics
1 answer:
nalin [4]2 years ago
5 0

The theoretical probability = \frac{1}{2}, The experimental probability  = \frac{3}{5}

Less than

<h3>What is experimental probability?</h3>

The experimental probability of an event occurring is the number of times that it occurred when the experiment was conducted as a fraction of the total number of times the experiment was conducted.

The theoretical probability of spinning and odd number = \frac{4}{8} = \frac{1}{2}

The experimental probability =\frac{6}{10} = \frac{3}{5}

Since  \frac{1}{2}<  \frac{3}{5}

So, the theoretical probability less than the experimental probability.

Learn more about experimental probability here:

brainly.com/question/3733849

#SPJ1

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Find the roots of the equation<br> x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits
scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

x^{2} + 3x - 8^{- 14} = 0

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

The above solution is for the quadratic equation of the form:

ax^{2} + bx + c = 0  

x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}

From the given eqn

a = 1

b = 3

c = - 8^{- 14}

Now, using the above values in the formula mentioned above:

x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)

Now, Rationalizing the above eqn:

x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}

x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}

Solving the above eqn:

x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}

Solving with the help of caculator:

x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}

The precise value upto three decimal places comes out to be:

x_{1, 1'} = 0.758\times 10^{- 14}

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