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3 years ago

Joshua buys tablets for $125. He marks them up to 35% to sell to his customers. What is the amount of mark-up?

Mathematics

1 answer:

koban [17]3 years ago

4 0

Answer:

The markup price of tablets is $168.75

Step-by-step explanation:

Given as :

The cost price of tablets = c.p = $125

The markup percentage = m = 35%

Let The markup amount = s.p

<u>According to question</u>

markup percentage = $\dfrac{\textr markup amount - cost price}{cost price}$

Or, m% = $\dfrac{s.p - c.p}{c.p}$

Or, 35% = $\dfrac{s.p-125}{125}$

Or, 125 × $\dfrac{35}{100}$ = s.p -125

Or, 125 × .35 + 125 = s.p

Or, s.p = 125 + 43.75

i.e s.p = $168.75

So, The markup price of tablets = s.p = $168.75

Hence, The markup price of tablets is $168.75 Answer

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

We can find the second moment given by:

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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:

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

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:

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

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