Emmett gathered some data to compare the monthly cost of renting a one-bedroom apartment in Dallas and Austin. The data collected for the monthly cost of renting a one-bedroom apartment in Dallas is shown below. $994 $1,322 $1,075 $1,189 $1,172
$1,465 $1,215 $930 $1,090 $1,288
The data collected for the monthly cost of renting a one-bedroom apartment in Austin is shown below.

$1,450 $1,400 $1,375 $1,389 $1,296
$900 $1,250 $1,100 $950 $1,495

Select ALL the statements that are true about the Austin Data.
This city's data had the higher interquartile range of the two cities by $87.
This city's data had the higher median monthly apartment cost of the two cities by $155.
this city's data had the lower average apartment cost because the mean was $86.50 less than that of the other set of data.
This city's data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data.

The correct options are;

a) The city's data has the higher median monthly apartment cost of the two cities by $155

b) The city's data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data

Step-by-step explanation:

The given data for the cost of renting a one-bedroom apartment in Dallas is presented as follows;

$994, $1,322, $1,075, $1,189, $1,172, $1,465, $1,215, $930, $1,090, $1,288

Which can be arranged in increasing order and analyzed to find the interquartile range, mean and standard deviation using Microsoft Excel as follows;

$930, $994, $1,075, $1,090, $1,172, $1,189, $1,215, $1,288, $1,322, and $ 1,465

The first quartile, Q₁ = $1,054.75

The third quartile, Q₃ = $1,296.5

The interquartile range = Q₃ - Q₁ = $241.75

The median = $1,180.5

The average monthly cost = $1,174

The standard deviation of the sample = $159.9597

The given data for the cost of renting a one-bedroom apartment in Austin is presented in increasing order using Microsoft Excel as follows;

$900, $950, $1,100, $1,250, $1,296, $1,375, $1,389, $1,400, $1,450, $1,495

The interquartile range, mean and standard deviation are found using Microsoft Excel as follows;

The first quartile, Q₁ = $1,062.5

The third quartile, Q₃ = $1,412.5

The interquartile range = Q₃ - Q₁ = $1,412.5 - $1,062.5 = $350

The median = $1,335.5

The average monthly cost = $1,260.5

The standard deviation of the sample = $209.0945

The difference in the median cost of the two cities is $1,335.5 - $1,180.5 = $155

Therefore, the Austin's city data has the higher median monthly apartment cost of the two cities by $155

b) The difference in the sample standard deviation of the two cities is $209.0945 - $159.9597 = $49.13476

Therefore, the Austin city data had the greater variability among monthly apartment costs because the standard deviation was $49.13 more than that of the other set of data.

a) $ 9.50

b) 17

c) y-intercept (0, 12). this is the initial amount of money that joe had prior to playing any game.

d) x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e) slope is -0.5. joe's balance will reduce by 0.5 for every game that he plays.

f)

domain; [0,24]

range; [0,12]

g) x + 2y = 24

step-by-step explanation:

a)

if joe plays 5 games, this will imply that x = 5. substitute x = 5 in the equation and solve for y;

y = -.50x + 12

y = -0.50(5) + 12

y = 9.50

b)

if joe leaves the arcade with $3.50, this will mean that y = 3.50. plug in y = 3.50 in the given equation and solve for x;

y = -.50x + 12

3.50 = -0.50x + 12

0.50x = 12 - 3.50

0.5x = 8.5

x = 17

c)

the function given is   y = -.50x + 12.

the y-intercept is the point where the graph of this function will intersect the y-axis. at this point, the value of x is usually 0. plug in x = 0 in the given equation and determine y;

y = -0.50(0) + 12

y = 12

the y-intercept is thus (0, 12). this is the initial amount of money that joe had prior to playing any game.

d)

the x-intercept refers to the point where the graph of a function intersects with the x-axis. at this point, the value of y is 0. substitute y = 0 in the given equation and solve for x;

0 =   -.50x + 12

0.5x = 12

x = 24

x-intercept ( 24, 0). this is the number of games that joe will play before he depletes the initial amount of money that he had.

e)

the slope of the given equation is -0.5. the equation has been given in slope-intercept form. thus the slope is simply the coefficient of x.

joe's balance will reduce by 0.5 for every game that he plays. this is the interpretation of the slope.

f)

domain is the set of x values for which the function is real and defined. the domain will thus be; [0,24]. this will restrict the balance to only assume positive values.

on the other hand, the range is the set of y values for which a function is real and defined. the range in our case will be [0,12]. joe can have between $0 and $12 balance.

g)

the standard form of the equation of a straight line is;

ax + by = c

where a, b, and c are integers.

our equation is given as;

y = -.50x + 12

re-writing the equation we have;

0.5x + y = 12

divide all through by 0.5;

x + 2y = 24

y > 1/3x

y < = -x + 4

hope this : )