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seraphim [82]
3 years ago
11

A multiple-choice test contains 10 questions. There are four possible answers for each question. a) In how many ways can a stude

nt answer the questions on the test if the student answers every question
Mathematics
1 answer:
AnnZ [28]3 years ago
3 0

Answer :

<h3>4^{10} <u> =1048576 ways </u> a student can answer the questions on the test if the student answers every question.</h3>

Step-by-step explanation:

Given that a multiple-choice test contains 10 questions and there are 4 possible answers for each question.

∴ Answers=4 options for each question.

<h3>To find how many ways  a student can answer the given questions on the test if the student answers every question :</h3>

Solving this by product rule

Product rule :

<u>If one event can occur in m ways and a second event occur in n ways, the number of ways of two events can occur in sequence is then m.n</u>

From the given the event of choosing the answer of each question having 4 options is given by

The 1st event of picking the answer of the 1st question=4 ,

2nd event of picking the answer of the 2nd question=4 ,

3rd event of picking the answer of the 3rd question=4

,....,

10th event of picking the answer of the 10th question=4.

It can be written as  by using the product rule

=4.4.4.4.4.4.4.4.4.4

=4^{10}

=1048576

<h3>∴ there are 1048576 ways a student can answer the questions on the test if the student answers every question.</h3>
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Answer:

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Step-by-step explanation:

To Find:

If discriminant (b^2 -4ac>0) how many real solutions

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Answer:

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(b) The standard deviation of <em>X</em> is 1.3.

Step-by-step explanation:

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The probability distribution of <em>X</em> id provided.

The formula to compute the mean or expected value of <em>X </em>is:

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The formula to compute the standard deviation of <em>X </em>is:

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E(X)=\sum x.P(X=x)\\=(0\times0.06)+(1\times0.25)+(2\times0.35)+(3\times0.15)+(4\times0.13)+(5\times0.06)\\=2.22\\\approx2.2

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(b)

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E(X^{2})=\sum x^{2}.P(X=x)\\=(0^{2}\times0.06)+(1^{2}\times0.25)+(2^{2}\times0.35)+(3^{2}\times0.15)+(4^{2}\times0.13)+(5^{2}\times0.06)\\=6.58

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\sigma=\sqrt{E(X^{2})-(E(X))^{2}}\\=\sqrt{6.58-(2.22)^{2}}\\=\sqrt{1.6516}\\=1.285\\\approx1.3

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