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mariarad [96]
2 years ago
4

B. Identify the kind of algebraic expression and determine the degree, variables

Mathematics
1 answer:
iren [92.7K]2 years ago
4 0

9514 1404 393

Answer:

  see the attachment

Step-by-step explanation:

There are no roots, fractional powers, or denominators containing variables, so each of these expressions is the sum of multiples of products of variables to integer powers. (One is a constant.) That makes each expression a <em>polynomial</em>. Some polynomials can be further classified by their degree.

The degree is the highest sum of powers of the variables in the terms. For example, the term a^2·b·c^2 has variables with powers 2, 1, and 2. The sum of those powers is 2+1+2 = 5, so this is a term of degree 5.

A constant is a term with no variables.

__

It is customary in algebra problems to use variables that are a single lower-case letter. Sometimes, a variable may be represented by a letter in uppercase, or in a special font or language (usually Greek). Sometimes, it may be a word or phrase. Here, we assume each variable is one letter, so xyz is the product of three variables.

Perhaps the table below answers the questions you are asking about these expressions.

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How do you solve -9( -4r + 6s ) + 3s - 7(6s - 2r)
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Answer: 2.1 Pull out like factors :

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Equation at the end of step

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Step-by-step explanation:

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The number of people that enter a drugstore in a given hour is a Poisson random variable with parameter λ = 10. Compute the cond
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Answer:

The probability is 0.2650

Step-by-step explanation:

Let's start assuming that men and women come in at the same rate.

Let's define the following random variables :

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M : ''Number of men that enter a drugstore''

W : ''Number of women that enter a drugstore''

The number of people will be the number of men plus the number of women

⇒

X = M + W

We are also assuming that M and W are independent random variables.

X ~ Po (10)

M ~ Po (λ1)

W ~ Po (λ2)

λ1 = λ2 because we assumed that men and women come in at the same rate.

λ1 = λ2 = λ

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We are looking for :

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