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How to solve algebraic equations with 2 variables

melisa1 [442]

You re write the equation, so when the equations are added one of the variables are eliminated

8 0

2 years ago

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz

Ahat [919]

Answer:

a) There is a 18.75% probability that the first question that she gets right is the second question.

b) There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

c) There is a 10.35% probability that she gets the majority of the questions right.

Step-by-step explanation:

Each question can have two outcomes. Either it is right, or it is wrong. So, for b) and c), we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem we have that:

Each question has 4 choices. So for each question, Robin has a $\frac{1}{4} = 0.25$ probability of getting ir right. So $\pi = 0.25$ . There are five questions, so $n = 5$ .

(a) What is the probability that the first question she gets right is the second question?

There is a 75% probability of getting the first question wrong and there is a 25% probability of getting the second question right. These probabilities are independent.

So

$P = 0.75(0.25) = 0.1875$

There is a 18.75% probability that the first question that she gets right is the second question.

(b) What is the probability that she gets exactly 1 or exactly 2 questions right?

This is: $P = P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 1) = C_{5,1}.(0.25)^{1}.(0.75)^{4} = 0.3955$

$P(X = 2) = C_{5,2}.(0.25)^{2}.(0.75)^{3} = 0.2637$

$P = P(X = 1) + P(X = 2) = 0.3955 + 0.2637 = 0.6592$

There is a 65.92% probability that she gets exactly 1 or exactly 2 questions right.

(c) What is the probability that she gets the majority of the questions right?

That is the probability that she gets 3, 4 or 5 questions right.

$P = P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 3) = C_{5,3}.(0.25)^{3}.(0.75)^{2} = 0.0879$

$P(X = 4) = C_{5,4}.(0.25)^{4}.(0.75)^{1} = 0.0146$

$P(X = 5) = C_{5,5}.(0.25)^{5}.(0.75)^{0} = 0.001$

$P = P(X = 3) + P(X = 4) + P(X = 5) = 0.0879 + 0.0146 + 0.001 = 0.1035$

There is a 10.35% probability that she gets the majority of the questions right.

6 0

2 years ago

Whats the answer to that question?

gavmur [86]

Answer:

-0.9090... can be written as $\frac{10}{11}$ .

Explanation:

Any repeating decimal can be written as a fraction by dividing the section of the pattern to be repeated by 9's.

We can start by listing out

0.909090... = 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...

Now. we let this series be equal to x, that is

$x$ = 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...

Now, we'll multiply both sides by 100 .

$100x$ = 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + ...

Then, subtract the 1st equation from the second like so:

$100x$ = 90 + 0 + 9/10 + 0/100 + 9/1000 + 0/10000 + 9/100000 + 0/1000000 + ...

$-x$ = - 9/10 - 0/100 - 9/1000 - 0/10000 - 9/100000 - 0/1000000 - ...

And we end up with this:

$99x=90$

Finally, we divide both sides by 99 in order to isolate x and get the fraction we're looking for.

$x=\frac{90}{99}$

Which can be reduced and simplified to

$x=\frac{10}{11}$

Hope this helps!

4 0

3 years ago

32 minus twice p is 31

Katena32 [7]

P=1 because 32-1 is 31

3 0

2 years ago

The theorems in this lesson relate to all of the following except for:

Katarina [22]

The theorems in this lesson relate to all of the following (tangents, arcs, and chords) except for radii.

7 0

2 years ago

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