An expression is shown below.
1 answer:
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The required proof is given in the table below:
![\begin{tabular}{|p{4cm}|p{6cm}|} Statement & Reason \\ [1ex] 1. $\overline{BD}$ bisects $\angle ABC$ & 1. Given \\ 2. \angle DBC\cong\angle ABD & 2. De(finition of angle bisector \\ 3. $\overline{AE}$||$\overline{BD}$ & 3. Given \\ 4. \angle AEB\cong\angle DBC & 4. Corresponding angles \\ 5. \angle AEB\cong\angle ABD & 5. Transitive property of equality \\ 6. \angle ABD\cong\angle BAE & 6. Alternate angles \end{tabular}](https://tex.z-dn.net/?f=%20%5Cbegin%7Btabular%7D%7B%7Cp%7B4cm%7D%7Cp%7B6cm%7D%7C%7D%20%0A%20Statement%20%26%20Reason%20%5C%5C%20%5B1ex%5D%20%0A1.%20%24%5Coverline%7BBD%7D%24%20bisects%20%24%5Cangle%20ABC%24%20%26%201.%20Given%20%5C%5C%0A2.%20%5Cangle%20DBC%5Ccong%5Cangle%20ABD%20%26%202.%20De%28finition%20of%20angle%20bisector%20%5C%5C%20%0A3.%20%24%5Coverline%7BAE%7D%24%7C%7C%24%5Coverline%7BBD%7D%24%20%26%203.%20Given%20%5C%5C%20%0A4.%20%5Cangle%20AEB%5Ccong%5Cangle%20DBC%20%26%204.%20Corresponding%20angles%20%5C%5C%0A5.%20%5Cangle%20AEB%5Ccong%5Cangle%20ABD%20%26%205.%20Transitive%20property%20of%20equality%20%5C%5C%20%0A6.%20%5Cangle%20ABD%5Ccong%5Cangle%20BAE%20%26%206.%20Alternate%20angles%0A%5Cend%7Btabular%7D)
![\begin{tabular}{|p{4cm}|p{6cm}|} 7. \angle AEB\cong\angle BAE & 7. Transitive property of equality \\ 8. \overline{EB}\cong\overline{AB} & 8. From 7. $\Delta ABE$ is isosceles \\ 9. EB = AB & 9. De(finition of congruence \\ 10. $\frac{AD}{DC}=\frac{EB}{BC}$ & 10. Triangle proportionality theorem \\ 11. $\frac{AD}{DC}=\frac{AB}{BC}$ & 11. Substitution Property of equality \\[1ex] \end{tabular} ](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%7B%7Cp%7B4cm%7D%7Cp%7B6cm%7D%7C%7D%0A7.%20%5Cangle%20AEB%5Ccong%5Cangle%20BAE%20%26%207.%20Transitive%20property%20of%20equality%20%5C%5C%0A8.%20%5Coverline%7BEB%7D%5Ccong%5Coverline%7BAB%7D%20%26%208.%20From%207.%20%24%5CDelta%20ABE%24%20is%20isosceles%20%5C%5C%0A9.%20EB%20%3D%20AB%20%26%209.%20De%28finition%20of%20congruence%20%5C%5C%2010.%20%24%5Cfrac%7BAD%7D%7BDC%7D%3D%5Cfrac%7BEB%7D%7BBC%7D%24%20%26%2010.%20Triangle%20proportionality%20theorem%20%5C%5C%0A11.%20%24%5Cfrac%7BAD%7D%7BDC%7D%3D%5Cfrac%7BAB%7D%7BBC%7D%24%20%26%2011.%20Substitution%20Property%20of%20equality%20%5C%5C%5B1ex%5D%20%0A%5Cend%7Btabular%7D%0A)
Answer:
P(X ≥ 1) = 0.50
Step-by-step explanation:
Given that:
The word "supercalifragilisticexpialidocious" has 34 letters in which 'i' appears 7 times in the word.
Then; the probability of success = 7/34 = 0.20588
Using Binomial distribution to determine the probability; we have:
where;
x = 0,1,2,...n and 0 < β < 1
and x represents the number of successes.
However; since the letter is drawn thrice; the probability that the letter "i" is drawn at least once can be computed as:
P(X ≥ 1) = 1 - P(X< 1)
P(X ≥ 1) = 1 - P(X =0)
P(X ≥ 1) = 1 - 0.50
P(X ≥ 1) = 0.50