Suppose that the grades of Business Mathematics and Statistic module is modeled well by a normal probability distribution with mean (212) and standard deviation (122). Let
X be the random variable representing this distribution. Find two symmetric values "a"
and "b" such that Probability [ a

a = -2.57 and b = 2.57

Step-by-step explanation:

Explanation:

Given mean of the Population = 212

Standard deviation of the Population = 122

Let  X be the random variable of the Normal distribution

Given P( a ≤ z ≤b) = 0.99

  put a=-b

         P( -b ≤ z ≤b) = 0.99

   ⇒     |A( b) - A( -b)| =0.99

    ⇒     | A( b) + A( b)| =0.99  

   ⇒       2 |A (b)| = 0.99

     ⇒     | A(b)| = 0.495

From normal table find value in areas

          b = 2.57 ( see in  normal table)

         Given a =-b

                   a = - 2.57

$18 was the price of the t-shirt

step-by-step explanation:

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.