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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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Sonbull [250]

Answer:

The positive value of k will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80, roots are those values of t so that h(t) = 0. That is:

19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0 (1)

Roots are determined analytically by the Quadratic Formula:

t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}

t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }

The smaller root is t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }, and the larger root is t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321}  }.

h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80 has one real root when \frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0. Then, we solve the discriminant for k:

\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}

k \approx \pm 0.028

The positive value of k will result in exactly one real root is approximately 0.028.

7 0
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Answer:
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Answer:

DOMAIN: -5, 2, 3 ,4

RANGE: 1, 3, 3 , 5

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Step-by-step explanation:

Since there is one value of y for every value of x this is a function.

The domain is just the x value (first number) and the range is just the y value(second number).

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Step-by-step explanation:

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-2=(2+x)/4

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