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3 years ago

12

In an examination there are four multiple choice questions with 3 options each, of which one option is the correct answer. In ho

w many ways can a student answer, such that he fails to answer all the four questions correctly? Thank you for help :)

1 answer:

mina [271]3 years ago

8 0

4 multiple choice questions
3 answers per question

4 × 3 = 12

<em>12 possible answers total</em>

12 - 4 = 8

<em>8 incorrect answers
</em><em>4 correct answers
</em><em>
</em>3 × 3 × 3 × 3 = 81
<em>
</em><em>81 possible ways they can answer
</em><em>
</em>Only one of those ways is 100% correct.
<em>
</em>81 - 1 = 80

80 ways

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Answer:

m = -8

Step-by-step explanation:

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3 years ago

Y=-x^2+2x+10<br> y=x+2<br><br> Substitution <br> Please show your work<br> Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

so

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

<em>Find the values of y</em>

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

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3 years ago

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kifflom [539]

Answer:

We need to conduct a hypothesis in order to determine if the mean is greater than specified value, the system of hypothesis would be:

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Alternative hypothesis: $\mu > \mu_o$

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$z_{crit}= 2.326$

Step-by-step explanation:

Notation

$\bar X$ represent the sample mean

$\sigma$ represent the standard deviation for the population

$n=$ sample size

$\mu_o$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean is greater than specified value, the system of hypothesis would be:

Null hypothesis: $\mu \leq \mu_o$

Alternative hypothesis: $\mu > \mu_o$

For this case the significance is 1%. So we need to find a critical value in the normal standard distribution who accumulates 0.99 of the area in the left and 0.01 in the right and for this case this critical value is:

$z_{crit}= 2.326$

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