Mathematics, 01.10.2019 21:30 angvtl02

Amultiple-choice test contains 25 questions, each with four answers. assume that a student just guesses on each question. (a) what is the probability that the student answers more than 20 questions correctly? (b) what is the probability that the student answers fewer than 5 questions correctly?

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Answer from: ramanpreet

(a) The probability of guessing 20 answers correctly is 1.15*10^-8 or 0.000001%

(b) The probability of answer fewer than 5 questions correctly is 0.2138 or 21.38%.

Step-by-step explanation:

If the student is guessing, he or she has only 1 out of 4 chances of answer correctly. This is equivalent to a probability of success p=0.25.

For this problem we can use the binomial distribution to calculate the probabilities.

(a) What is the probability that the student answers more than 20 questions correctly?

We have a total of 25 questions. The probability of guessing 20 answers correctly is

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}\\\\P(20)=\frac{25!}{20!5!}*0.25^{20}*0.75^5\\\\P(20)=53130* 9.095*10^{-3}*0.2373\\\\P(20)=1.15*10^{-8}$

The probability of guessing 20 answers correctly is 1.15*10^-8 or 0.000001% .

(b) The probability of answer fewer than 5 questions correctly is the sum of:

1) Answering 0 questions correctly

2) Answering 1 questions correctly

3) Answering 2 questions correctly

4) Answering 3 questions correctly

5) Answering 4 questions correctly

This probabilities can be calculated as

$P(0)=\frac{25!}{0!25!}*0.25^0*0.75^{25}=1*1*7.53*10^{-4}=0.0008\\\\P(1)=\frac{25!}{1!24!}*0.25^1*0.75^{24}= 0.0063 \\\\P(2)=\frac{25!}{2!23!}*0.25^2*0.75^{23}= 0.0251 \\\\P(3)=\frac{25!}{3!22!}*0.25^3*0.75^{22}= 0.0641 \\\\P(4)=\frac{25!}{4!21!}*0.25^4*0.75^{21}= 0.1175$

Then the sum of the probabilty of this events is

P(x

The probability of answer fewer than 5 questions correctly is 0.2138 or 21.38%.

Answer from: Quest

143 75 107 13 the answer
Enter the missing numbers in the boxes to complete the table of equivalent ratios of lengths to widt

Answer from: Quest

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

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