The median for the last 20 seasons was 7 wins. Arranged in order from least to greatest, the middle two seasons had wins of 6 games and 8 games. In the 21st season, the team won 14 games. What will be the new median? Show or explain your thinking.

8 wins

Step-by-step explanation:

Median is the middle number of a set of numbers or data set.

For a set of twenty numbers, the middle two seasons are the 10th and 11th seasons and the median is given by the mean of both numbers.

10th season = 6 wins

11th season = 8 wins

i.e (6+8)/2 = 14/2 = 7 wins

Now, the number of wins for the 21st season is added to this data set, the median number for 21 seasons (numbers) will be the 11th season (number).

Following the arrangement of numbers by seasons, the middle season is the 11th season and the 11th season has 8 wins.

Hence, the new median win is 8 wins.

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841

answer

you start with $5 and save $10 a week until you have at least $45wer

step-by-step explanation: