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3 years ago

Which is the better buy, 3 pounds for $8.31 or 5 pounds for 12.59

1 answer:

3 0

If 3 pounds cost $8.31, then the product is being sold as 1 for $2.77.

If 5 pounds cost $12.59, then the product is being sold as 1 for $2.51.

It is better to buy 5 pounds for $12.59 because you are getting more for your money.

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Need help on my math work here is a pic of the question

faltersainse [42]

Answer:

-3

Step-by-step explanation:

A = midpoint

-17 + 5 = -12

-12/2 = -6 = A

-6 + 0 = -6

-6/2 = -3 = B

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Solve this 25 points

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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

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:

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

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3 years ago

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alex41 [277]

Answer:

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Step-by-step explanation:

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i think i found it, but you might have to print it out, but i will send the link: https://www.murrieta.k12.ca.us/cms/lib5/CA01000508/Centricity/Domain/1791/_Files/Algebra%201/c10rev.pdf

:) have a nice day love ❤

Step-by-step explanation:

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