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kolbaska11 [484]
2 years ago
8

Arrange the reasons for the proof in the correct order. Prove: If the diameter of a circle is 6 meters and the formula for diame

ter is d = 2r, then the radius of the circle is 3 meters.
Mathematics
1 answer:
Misha Larkins [42]2 years ago
3 0
Given the question: Arrange the reasons for the proof in the correct order. Prove: If the diameter of a circle is 6 meters and the formula for diameter is d = 2r, then the radius of the circle is 3 meters.
A. If r ≠ 3 m, then d ≠ 6 m. Since the contrapositive is true, the original statement must also be true. Therefore, if the diameter of a circle is 6 meters, then the radius is 3 meters.
B. Multiplication of real numbers shows that d = 2(2 m) = 4 m.
C. Substitute r = 2 m into d = 2r.
D. Assume that r ≠ 3 m. For example, the radius equals another length, such as r = 2 m.

To prove that i<span>f the diameter of a circle is 6 meters and the formula for diameter is d = 2r, then the radius of the circle is 3 meters by contradiction, we assume that the radius in not equal to 3 meters, for </span>example, the radius equals another length, such as r = 2 m.
Next, we substitute the value: r = 2m nto the original equation that says that d = 2r, i.e. d = 2(2m) = 4m which is not true and contradicts the original statement that the diameter of the circle is 6m.

Therefore, the arrangement of the proof is as follows:
 D. Assume that r ≠ 3 m. For example, the radius equals another length, such as r = 2 m.
C. Substitute r = 2 m into d = 2r.
B. Multiplication of real numbers shows that d = 2(2 m) = 4 m.
A. If r ≠ 3 m, then d ≠ 6 m. Since the contrapositive is true, the original statement must also be true. Therefore, if the diameter of a circle is 6 meters, then the radius is 3 meters.

D C B A.
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The subsets of the real numbers can be represented in a Venn diagram as shown. To which subset of real numbers does −2π belong?
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D

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Let Xn be the random variable that equals the number of tails minus the number of heads when n fair coins are flipped. What is t
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the expected value of Xn , E(Xn) = 0 and the variance σ²(Xn) = n*(1-2n)

Step-by-step explanation:

If X1= number of tails when n fair coins are flipped , then X1 follows a binomial distribution with E(X1) = n*p , p=0,5 and the number of heads obtained is X2=n-X1

therefore

Xn =X1-X2 = X1- (n-X1) = 2X1-n

thus

E(Xn) =∑ (2*X1-n) p(X1) =  2*∑[X1 p(X1)] -n∑p(X1) = 2*E(X1)-n = 2*n*p--n= 2*n*1/2 -n = n-n =0

the variance will be

σ²(Xn) = ∑ [Xn - E(Xn)]² p(Xn) = ∑ [(2X1-n) - 0 ]² p(X1) = ∑ (4*X1²-4*X1*n+n²) p(X1) = = 4*∑ X1²p(X1) - 4n ∑X1 p(X1) -  n²∑p(X1) = 2*E(X1²) -4n*E(X1)- n²

since

σ²(X1) = n*p*(1-p) = n*0,5*0,5=n/4

and

σ²(X1) = E(X1²) - [E(X1)]²

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E(X1²) = n(n+1)/4

therefore

σ²(Xn) = 4*E(X1²) -4n*E(X1)- n² = 4*n(n+1)/4 - 4*n*n/2 - n² = n(n+1) - 2n² - n²

= n - 2n² = n(1-2n)

σ²(Xn) = n(1-2n)

4 0
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