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Karolina [17]
3 years ago
11

Stacy uses a spinner with six equal sections numbered 2, 2, 3, 4, 5, and 6 to play a game. Stacy spins the pointer 120 times and

records the results. The pointer lands 30 times on a section numbered 2, 19 times on 3, 25 times on 4, 29 times on 5, and 17 times on 6. How does the theoretical probability compare with the experimental probability of landing on an even number? Use the drop-down menus to explain your answer.
The probability is greater than the probability of landing on an even number. The probability is 72120 and the probability is 80120.
Mathematics
1 answer:
ivann1987 [24]3 years ago
5 0

Answer:

Even numbers are 2,4,6 for this case and the experimental probability of being even is given by:

p = \frac{30+25+17}{120}= \frac{72}{120}= 0.6

Now we can find the theorical probability since we have 6 possible outcomes 2, 2, 3, 4, 5, and 6 we have 4 even numbers in the sample space and the theorical probability is given by:

p= \frac{4}{6}= 0.667

Then we can conclude that the empirical probability is lower than the theorical probability (0.6<0.667)  for this experiment

Step-by-step explanation:

For this case we wan find the probability associated to the experiment we know that. The pointer lands 30 times on a section numbered 2, 19 times on 3, 25 times on 4, 29 times on 5, and 17 times on 6.

Even numbers are 2,4,6 for this case and the experimental probability of being even is given by:

p = \frac{30+25+17}{120}= \frac{72}{120}= 0.6

Now we can find the theorical probability since we have 6 possible outcomes 2, 2, 3, 4, 5, and 6 we have 4 even numbers in the sample space and the theorical probability is given by:

p= \frac{4}{6}= 0.667

Then we can conclude that the empirical probability is lower than the theorical probability (0.6<0.667)  for this experiment

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B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt
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B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2

Let's assume 0, so that |\csc\theta|=\csc\theta.

Now,

\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}
\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}
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Step-by-step explanation:

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The green triangle is a dilation of the red triangle with a scale factor of s=13 and the center of dilation is at the point (4,2
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Given:

The red figure dilated with a scale factor of s=\dfrac{1}{3} and the center of dilation is at the point (4,2) to get the green figure.

To find:

The coordinates of C' and A.

Solution:

If a figure is dilated with a scale factor k and the center of dilation is at the point (a,b), then

(x,y)\to (k(x-a)+a,k(y-b)+b)

In given problem, the scale factor is \dfrac{1}{3} and the center of dilation is at (4,2).

(x,y)\to (\dfrac{1}{3}(x-4)+4,\dfrac{1}{3}(y-2)+2)            ...(i)

Let the vertices of red triangle are A(m,n), B(10,14) and C(-2,11).

Using (i), we get

C(-2,11)\to C'(\dfrac{1}{3}(-2-4)+4,\dfrac{1}{3}(11-2)+2)

C(-2,11)\to C'(\dfrac{1}{3}(-6)+4,\dfrac{1}{3}(9)+2)

C(-2,11)\to C'(-2+4,3+2)

C(-2,11)\to C'(2,5)

Therefore, the coordinates of Point C' are C'(2,5).

We assumed that point A is A(m,n).

Using (i), we get

A(m,n)\to A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)

From the given figure it is clear that the image of point A is (8,4).

A'(\dfrac{1}{3}(m-4)+4,\dfrac{1}{3}(n-2)+2)=A'(8,4)

On comparing both sides, we get

\dfrac{1}{3}(m-4)+4=8

\dfrac{1}{3}(m-4)=8-4

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(n-2)=3(2)

n=6+2

n=8

Therefore, the coordinates of point A are (16,8).

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