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3 years ago

8

On a piece of paper, use a protractor and a ruler to construct △ABC with AB=4 in. , AC=4 in. , and m∠A=60° . Which statement is

true about this triangle?
A) BC=3 in.
B) BC=4 in.
C) BC=5 in.
D) BC can not be measured

2 answers:

Fed [463]3 years ago

5 0

Answer: B) BC=4 in.

Step-by-step explanation:

Given: In △ABC , AB=4 inch ,AC=4 in. , and m∠A=60°

⇒ AB=AC= 4 inches

⇒ △ABC is an isosceles triangle.

⇒ m∠A=60°=m∠B [angles opposite to the equal sides of a triangle are equal]

Now by angle sum property,

m∠A+m∠B+m∠C=180°

⇒60°+60°+m∠C=180°

⇒ m∠C=180°-120°

⇒ m∠C=60°

⇒ △ABC is an equilateral triangle.

⇒ BC= 4 inches [sides of an equilateral triangles are equal]

⇒ B is the right answer.

Likurg_2 [28]3 years ago

5 0

Answer: B) BC=4 in.

Step-by-step explanation:

Given: In △ABC , AB=4 inch ,AC=4 in. , and m∠A=60°

⇒ AB=AC= 4 inches

⇒ △ABC is an isosceles triangle.

⇒ m∠A=60°=m∠B [angles opposite to the equal sides of a triangle are equal]

Now by angle sum property,

m∠A+m∠B+m∠C=180°

⇒60°+60°+m∠C=180°

⇒ m∠C=180°-120°

⇒ m∠C=60°

⇒ △ABC is an equilateral triangle.

⇒ BC= 4 inches [sides of an equilateral triangles are equal]

⇒ B is the right answer.

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$E(x)=-0.2101$

Step-by-step explanation:

The expected value for a discrete variable is calculated as:

$E(x)=x_1*p(x_1)+x_2*p(x_2)$

Where $x_1$ and $x_2$ are the values that the variable can take and $p(x_1)$ and $p(x_2)$ are their respective probabilities.

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If there are n identical and independent events with a probability p of success and a probability (1-p) of fail, the probability that a events from the n are success are equal to:

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Where $nCa=\frac{n!}{a!(n-a)!}$

So, in this case, n is number of times that the player tosses a die and p is the probability to get a 6. n is equal to 6 and p is equal to 1/6.

Therefore, the probability $p(x_1)$ that a player get at least two times number 6, is calculated as:

$p(x_1)=P(x\geq2)=1-P(0)-P(1) \\\\P(0) =6C0*(1/6)^{0}*(5/6)^{6}=0.3349\\P(1)=6C1*(1/6)^{1}*(5/6)^{5}=0.4018\\\\p(x_1)=1-0.3349-0.4018\\p(x_1)=0.2633$

On the other hand, the probability $p(x_2)$ that a player don't get at least two times number 6, is calculated as:

$p(x_2)=1-p(x_1)=1-0.2633=0.7367$

Finally, the expected value of the amount that the player wins is:

$E(x)=x_1*p(x_1)+x_2*p(x_2)\\E(x)=2*(0.2633)+ (-1)*0.7367\\E(x)=-0.2101$

It means that he can expect to loses 0.2101 dollars.

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