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DIA
1 month ago
7

In the spinner below, the large wedges are twice the size of the smaller ones. What is true about the probability of landing on

2 and the probability of landing on 3?
A.The probability of landing on 2 is half the probability of landing on 3.
B.The probabilities cannot be predicted.
C.The probabilities are equal.
D.The probability of landing on 2 is twice the probability of landing on 3.

Mathematics
2 answers:
tester [12.3K]1 month ago
6 0

Response:

C

Detailed explanation:

I would choose option c as the spinner has a chance to land on any area at any moment. I wish I could provide a definitive answer.

tester [12.3K]1 month ago
4 0

Response:

The chance of landing on 2 is twice that of landing on 3.

Detailed explanation:

This is because the section associated with 2 is twice the size of the section associated with 3, indicating that the likelihood of landing on 2 is double that of landing on 3.

You might be interested in
A straight rod has one end at the origin and the other end at the point (l,0) and a linear density given by λ=ax2, where a is a
Svet_ta [12734]

It is stated that a straight rod has one endpoint at the origin (0,0) and the opposite endpoint at (L,0), with a linear density defined by \lambda=ax^2, where a is a constant and x is the x coordinate.

Thus, the infinitesimal mass is expressed as:

dm=\lambda \times dx=\lambda dx

The total mass can be calculated by integrating the above expression as follows:

\int\,dm= \int\limits^L_0 {ax^2} \, dx

Consequently, m=a\int\limits^L_0 {x^2} \, dx=a[\frac{x^3}{3}]_{0}^{L}=\frac{a}{3}[L^3-0]= \frac{aL^3}{3}

Now, we can calculate the center of mass, x_{cm} of the rod as:

x_{cm}=\frac{1}{m} \int xdm

x_{cm}=\frac{1}{m}\int_{0}^{L}x\times \lambda dx =\int_{0}^{L}x\times ax^2 dx=\int_{0}^{L}ax^3 dx

Now, it follows that

x_{cm}=\frac{1}{\frac{aL^3}{3}}\int_{0}^{L}ax^3dx=\frac{3}{aL^3}\times [\frac{ax^4}{4}]_{0}^{L}

Therefore, the center of mass, x_{cm} is located at:

\frac{3}{aL^3}\times [\frac{ax^4}{4}]_{0}^{L}=\frac{3}{aL^3}\times \frac{aL^4}{4}=\frac{3}{4}L


5 0
2 months ago
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