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4 years ago

2. Two angles have a common vertex but no common side. How can you tell whether the angles are vertical angles?

Mathematics

1 answer:

Maksim231197 [3]4 years ago

7 0

Explanation:

The angles are <em>vertical angles</em> if the opposites of the rays forming one of the angles are the rays forming the other angle.

More formally, if V is the common vertex, and ...

R is a point on one of the rays forming Angle 1
S is a point on the ray that is the opposite of ray VR
T is a point on the other ray forming Angle 1
U is a point on the ray that is the opposite of ray VT

Then angle RVT and angle SVU are vertical angles.

___

Another way to say this is that points R, V, S are collinear, as are points T, V, U, and the two angles of interest are RVT and SVU.

If the above conditions cannot be met, then the angles are not vertical angles.

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3 years ago

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denis23 [38]

Answer:

(1) The possible outcomes are: X = {0, 1, 2, 3}.

(2) The number of times should Hartley spin a difference of 1 is 36.

(3) The number of times should Hartley spin a difference of 0 is 24.

Step-by-step explanation:

The number of sections on the spinner is 4 labelled as {1, 2, 3, 4}.

The total number of spins for each of the spinner is, <em>n</em> = 96.

(1)

The sample space of spinning both the spinners together are:

S = {(1, 1), (1, 2), (1, 3), (1, 4)

(2, 1), (2, 2), (2, 3), (2, 4)

(3, 1), (3, 2), (3, 3), (3, 4)

(4, 1), (4, 2), (4, 3), (4, 4)}

Total = 16.

The possible outcomes are:

X = {0, 1, 2, 3}.

(2)

The sample space with the difference 1 are:

S₁ = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3)}

n (S₁) = 6

The probability of the difference 1 is:

$P(\text{Diff}=1)=\frac{n(S_{1})}{N}=\frac{6}{16}=\frac{3}{8}$

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 1 is:

$E(\text{Diff}=1)=P(\text{Diff}=1)\times n\\\\=\frac{3}{8}\times 96\\\\=36$

Thus, the number of times should Hartley spin a difference of 1 is 36.

(3)

The sample space with the difference 0 are:

S₂ = {(1, 1), (2, 2), (3, 3), (4, 4)}

n (S₂) = 4

The probability of the difference 0 is:

$P(\text{Diff}=0)=\frac{n(S_{2})}{N}=\frac{4}{16}=\frac{1}{4}$

The spinners were spinner 96 times.

The expected number of times would Hartley spin a difference of 0 is:

$E(\text{Diff}=0)=P(\text{Diff}=0)\times n\\\\=\frac{1}{4}\times 96\\\\=24$

Thus, the number of times should Hartley spin a difference of 0 is 24.

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3 years ago