Problem is on the attached picture. Thanks!

Cost of ribeye steak dinner is $19.058

Cost of grilled salmon dinner is $14.88

Solution:

Let x be the ribeye steak dinner and y be the grilled salmon dinner.

Write the given equation in the algebraic form.

First day sold:

15x + 21 y = 598.34 – – – – (1)

Second day sold:

25x + 7y = 580.61 – – – – (2)

To solve these equations by elimination method.

Multiply equation (2) by 3, we get

75x + 21y = 1741.83 – – – – (3)

Now, subtract equation (1) from equation (3).

75x + 21y – (15x + 21y)= 1741.83 – 598.34

75x + 21y – 15x – 21y = 1143.49

60x = 1143.49

Divide by 60 on both sides of the equation.

x = 19.058

Substitute x value in equation (2), we get

25(19.058) + 7y = 580.61

476.45 + 7y = 580.61

Subtract 476.45 on  both sides of the equation.

7y = 104.16

Divide by 7 on both sides of the equation.

y = 14.88

Cost of ribeye steak dinner is $19.058

Cost of grilled salmon dinner is $14.88.

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

6.   b - there is a high data value that causes the data set to be asymmetrical for the males: the data for males are high and asymmetrical .   there is an outlier for the males that skews the data

5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.