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OverLord2011 [107]
2 years ago
11

The admission fee at a small fair is $1.50 for children and $4.00 for adults. On a certain day, 2,200 people enter the fair and

$5,050 is collected. How many children and how many adults attended?
Mathematics
1 answer:
trapecia [35]2 years ago
8 0

Given :

The admission fee at a small fair is $1.50 for children and $4.00 for adults.

On a certain day, 2,200 people enter the fair and $5,050 is collected.

To Find :

How many children and how many adults attended.

Solution :

Let, number of children and adults attended are c and a.

So, c + a = 2200      ....1)

Now, fair collected is given by :

1.5c + 4a = 5050     ....2)

Putting value of a from equation 1) to 2), we get :

1.5c + 4( 2200 - c ) = 5050

1.5c - 4c = 5050 - 8800

2.5c = 3750

c = 1500

a = 2200 - 1500

a = 700

Therefore, 1500 children and 700 adults attended the fair.

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Write the equation of the line that is parallel to y=-0.75x and that passes through the point (8,0)
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Equation of line parallel to y = -0.75x that passes through (8,0) is: y = -0.75x-6

Step-by-step explanation:

Given equation of line is:

y=-0.75x

The coefficient of x is the slope of the line so the slope of given line is: -0.75

As parallel lines have equal slopes so the required equation of line will have the same slope.

The slope-intercept form is:

y = mx+b

Putting the value of the slope

y = -0.75x+b

To find the value of b, putting the point in the equation

0 = -0.75(8)+b\\0 = -6+b\\b = 6

Putting the values of m and b, we get

y = -0.75x-6

Hence,

Equation of line parallel to y = -0.75x that passes through (8,0) is: y = -0.75x-6

Keywords: Equation of line, Slope-intercept form

Learn more about equation of line at:

  • brainly.com/question/10435816
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3 years ago
Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviat
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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

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

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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

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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

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