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Liula [17]
3 years ago
5

The figure below shows segments AC and EF which intersect at point B. Segment AF is parallel to segment EC.. . Which of these fa

cts is used to prove that triangle ABF is similar to triangle CBE?. . Angle AFB is congruent to angle CEB because corresponding angles are congruent.. . Angle AFB is congruent to angle CEB because corresponding angles are congruent.. . Line segment AF is congruent to line segment EC because parallel segments are congruent.. . Line segment AB is congruent to line segment FB because legs of an isosceles triangle are congruent.
Mathematics
1 answer:
sertanlavr [38]3 years ago
8 0
I saw the image that should have accompanied this problem.

Two triangles were formed, one smaller than the other but it can be said that both triangles are similar with one another. 

<span>The fact used to prove that triangle ABF is similar to triangle CBE is: 

Angle AFB is congruent to angle CEB because corresponding angles are congruent</span>
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Numbers with the same absolute value are
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✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ The correct answer is D. equally distant

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~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

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The probability that an American CEO can transact business in a foreign language is .20. Ten American CEOs are chosen at random.
elena-14-01-66 [18.8K]

Answer:

(a) 0.1074

(b) 0.6342

(c) 1.024 x 10⁻⁷

Step-by-step explanation:

This problem can be modeled as a binomial probability with probability of success (transacting business in a foreign language) p =0.20, and n = 10 trials. The binomial probability model is:

P(X=x) = \frac{n!}{(n-x)!x!}*p^x*(1-p)^{n-x}

(a) None can transact business in a foreign language, P(X=0)

P(X=0) = \frac{10!}{(10-0)!0!}*0.2^0*(1-0.2)^{10-0}\\P(X=0) = 0.8^{10} = 0.1074

(b) At least two can transact business in a foreign language

P(X\geq2) = 1-(P(X=0)+P(X=1))\\P(X\geq2) = 1-(0.1074+\frac{10!}{(10-1)!1!}*0.2^1*(1-0.2)^{10-1})\\P(X\geq2) = 1- (0.1074+10*0.2*0.8^{9}) = 0.6342

(c) All 10 can transact business in a foreign language

P(X=10) = \frac{10!}{(10-10)!10!}*0.2^{10}*(1-0.2)^{10-10}\\P(X=10) = 0.2^{10} = 1.024 * 10^{-7}

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3 years ago
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