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Montano1993 [528]
3 years ago
5

PLEASE HELP NEEDS TO BE DONE TONIGHT HELP PLEASE ASAP!

Mathematics
1 answer:
Levart [38]3 years ago
8 0

Answer:

<em></em>

  • <em>    Variable</em>
  • <em>    Variable expression</em>
  • <em>    Number</em>

Explanation:

A <em>literal equation</em> contains numbers, letters, and signs.

The letters represent variables, the numbers are constants, and the sign express relations, like equal (=), less than (<), and greater than (>).

You use the signs to make equations or inequalities.

Algebra is the use of mathematical rules applied to literal expressions. You can add, subtract, multiply and divide literal, expressions, like in arithmetic,  following some rules.

<em>To rewritte a literal equation in terms of a specific variable</em>, you may need to <em>divide by</em>  <em>another variable, a variable expression, or a number</em>, but never by an inequality sign.

Here you have one example for each case:

<u>1. Dividing by another variable.</u>

     x^2y^5=100\\\\\\ \dfrac{x^2y^5}{y^5}=\dfrac{100}{y^5}\\ \\ \\ x^2=\dfrac{100}{y^5}

<u>  2. Dividing by a variable expression.</u>

      x^2(z^3+y^5)=100\\\\\\ \dfrac{x^2(z^3+y^5)}{(z^3+y^5)}=\dfrac{100}{(z^3+y^5)}\\ \\ \\ x^2=\dfrac{100}{(z^3+y^5)}

<u>3. Dividing by a number:</u>

   

      200x^2=100\\\\\\ \dfrac{200x^2}{200}=\dfrac{100}{200}\\ \\ \\ x^2=\dfrac{100}{200}\\ \\ \\ x^2=0.5

As said above,  you cannot divide by an inequality sign.

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Answer:

It is very unusual as the probability to get a sample average of 75 or more customers if the manager had not offered the discount is 0.006    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 50

Standard Deviation, σ = 10

We are given that the distribution of number of customers is a bell shaped distribution that is a normal distribution.

In an attempt to increase the number of weekday customers, the manager offers a $10 membership discount on 5 consecutive weekdays and  the sample mean of customers during this weekday period jumps to 75.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P(sample average of 75 or more customers if the manager had not offered the discount)

P(x > 75)

P( x > 75) = P( z > \displaystyle\frac{75 - 50}{10}) = P(z > 2.5)

= 1 - P(z \leq 2.5)

Calculation the value from standard normal z table, we have,  

P(x > 610) = 1 - 0.994 = 0.006= 0.6\%

It is very unusual as the probability to get a sample average of 75 or more customers if the manager had not offered the discount is 0.006

b) Yes, the managers discount strategy worked as it increased the average number of customers from 50 customers to 75 customers.

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