59 can't go into 9, but if u mean the other way then it is 9.83
Answer:
y
Step-by-step explanation:
The variable in 8y^3 is y because a variable is a number that we don't know yet, and that is the only one.
Answer:
or 
Step-by-step explanation:
Let
h ----> a skier's height in centimeters
we know that
A ski shop sells skis with lengths ranging from 150 centimeters to 220 centimeters
The interval of the skis is -----> [150,220]
The compound inequality that represents the heights (in centimeters) of skiers the shop does not provide skis is
----> inequality A
or
----> inequality B
<em>Solve inequality A</em>

Divide by 1.16 both sides

<em>Solve inequality B</em>

Divide by 1.16 both sides

therefore
or 
Answer:
Check the explanation
Step-by-step explanation:
(a) The appropriate test is the matched-pairs test because a student’s score on Try 1 is certainly correlated with his/her score on Try 2. Using the differences, we have xbar = 29 and s = 59.
(b) To test H0: mu=0 vs. H1 mu > 0, we compute

with df = 426. This is certainly significant, with P < 0.0005. Coached students do improve their scores on average
(a) H0: μ1 = μ2 vs. Ha: μ1 > μ2, where μ1 is the mean gain among all coached students and μ2 is the mean gain among uncoached students. H0 and Ha. Using the conservative approach, df = 426 is rounded down to df = 100 in (t table) and we obtain 0.0025 < P < 0.005. Using software, df = 534.45 and P = 0.004. There is evidence that coached students had a greater average increase.
(b) 8 ± t*(3.0235) where t* equals 2.626 (using df = 100 with (t table) ) or 2.585 (df = 534.45 with software). This gives either 0.06 to 15.94 points, or 0.184 to 15.816 points, respectively.
(c) Increasing one’s score by 0 to 16 points is not likely to make a difference in being granted admission or scholarships from any colleges.
Answer:
11/30, 2/5, 7/15
Step-by-step explanation: