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3 years ago

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A toy store buys action figures from a factory for $8.00 per action figure. If they mark-up the toys 110% before selling them, w

hat is the price of one of the action figures in the store?

2 answers:

grin007 [14]3 years ago

6 0

Answer:

<em>$</em><em>8.80</em>

Step-by-step explanation:

<em>1</em><em>1</em><em>0</em><em>%</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>whole</em><em> </em><em>cost</em><em> </em><em>plus</em><em> </em><em>1</em><em>0</em><em>%</em><em>.</em><em> </em><em>you</em><em> </em><em>have</em><em> </em><em>you're</em><em> </em><em>cost</em><em> </em><em>at</em><em> </em><em>8</em><em>$</em><em> </em><em>add</em><em> </em><em>1</em><em>0</em><em>%</em><em> </em><em>and</em><em> </em><em>that</em><em> </em><em>is</em><em> </em><em>80</em><em> </em><em>cents</em><em>.</em>

polet [3.4K]3 years ago

4 0

Answer:

$8.80

Step-by-step explanation:

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Select the statement that is FALSE. The range is never greater than the greatest value of a data set. The interquartile range is

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Answer:

The interquartile range is the difference between the highest and lowest values in the middle of a data set.

Step-by-step explanation:

The range is the difference between the maximum and minimum value, hence, it cannot be greater than the maximum value, which is the greatest value in a dataset, the highest value a range could have being equal to the maximum value when the minimum vlaue of the dataset is equal to 0.

The mean is the average value of a dataset, hence, it cannot be greater than the maximum value.

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3 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.

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\hbox{solve the 2nd equation for y:} \\
-7x+y=-23 \\
y=7x-23 \\ \\
\hbox{substitute 7x-23 for y in the 1st equation:} \\
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\boxed{\hbox{infinitely many solutions}}$

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Answer:

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