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Nimfa-mama [501]
3 years ago
14

The teachers’ edition of a statistics textbook sells for $150, and students’ edition of the book sell for $50 each. which functi

on can be used to find the average cost per book if two teachers’ editions and x students’ editions are purchased?
a. f(x)= 150 50x over 1 x
b. f(x)= 150 50x over 2 x
c. f(x)= 300 50x over 1 x
d. f(x)= 300 50x over 2 x
Mathematics
2 answers:
natima [27]3 years ago
4 0
Its d) f(x)= 300+50x/2+x..............................................................................
Juliette [100K]3 years ago
4 0

Answer:

d

Step-by-step explanation:

Lets have a look at what is given:

Teachers edition of statistics textbook sells for $150

Students edition of statistics textbook sells for $50.

Two teachers and x student books are purchased.

Let f(x) = the average cost per book

Therefore the amount of the teachers textbook cost 2*$150 = $300

and the students texbook costs x*$50= $50x

The average cost per book is:

f(x) = (300+50x)/(x+2)

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Answer:

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Which box-and-whisker plot represents the data set?<br><br> 11, 6, 19, 14, 21, 7, 13, 15, 15
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The box and whisker plot is attached.

We first order the data from least to greatest:
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We draw the middle line of the box at 14, the median.  We draw the left side of the box at the lower quartile, 9.  We draw the right side of the box at the upper quartile, 17.  From the right side of the box, we draw a whisker to the highest value, 21. From the left side of the box, we draw a whisker to the lowest value, 6.

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Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat
drek231 [11]

Answer:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

Step-by-step explanation:

For this case we have the following probability distribution given:

X          0            1        2         3        4         5

P(X)   0.031   0.156  0.313  0.313  0.156  0.031

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

We can verify that:

\sum_{i=1}^n P(X_i) = 1

And P(X_i) \geq 0, \forall x_i

So then we have a probability distribution

We can calculate the expected value with the following formula:

E(X) = \sum_{i=1}^n X_i P(X_i) = 0*0.031 +1*0.156+ 2*0.313+3*0.313+ 4*0.156+ 5*0.031 = 2.5

We can find the second moment given by:

E(X^2) = \sum_{i=1}^n X^2_i P(X_i) = 0^2*0.031 +1^2*0.156+ 2^2*0.313+3^2*0.313+ 4^2*0.156+ 5^2*0.031 =7.496

And we can calculate the variance with this formula:

Var(X) =E(X^2) -[E(X)]^2 = 7.496 -(2.5)^2 = 1.246

And the deviation is:

Sd(X) = \sqrt{1.246}= 1.116

6 0
3 years ago
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