Answer:
Distance swam by Sonya in 3 weeks = 19740 m
Step-by-step explanation:
To practice for a competition, Sonya swam 0.94 kilometer in the pool each day for 3 weeks.
Number of days in 1 week = 7
Number of weeks she swam = 3
Total number of days she swam = 3 x 7 = 21 days
Distance swam per day = 0.94 km
Total distance swam = Total number of days she swam x Distance swam per day
Total distance swam = 21 x 0.94 = 19.74 km = 19.74 x 1000 m = 19740 m
Distance swam by Sonya in 3 weeks = 19740 m
Answer: the answer is b( a quantile - quantile plot and a p- value test
Step-by-step explanation:
Answer:
top row on 9 page; 9) 53/5 10) 26/4 11) 37/4
bottom row on 9 page; 9) 8 and 1/7 10) 6 and 3/4 11) 1 and 1/3
top row on 3 page; 3) 2 and 2/7 4) 5 and 3/4 5) 8 and 1/10
bottom row on 3 page; 3) 4/3 4) 3/2 5) 12/5
top row on 12 page; 12) 21/10 13) 62/6 14) 57/6
bottom row on 12 page; 12) 1 and 9/10 13) 10 and 1/2 14) 3 and 3/8
Step-by-step explanation:
You never specified if these had to be simplified or turned into a fraction, so I just simplified them. That's about it.
I hope this helps :)
Answer:
4/5 x 7/12 = 28/60 as simplified as 7/15 in fraction form. 4/5 x 7/12 = 0.4667 in decimal form.
Answer:
The difference in the sample proportions is not statistically significant at 0.05 significance level.
Step-by-step explanation:
Significance level is missing, it is α=0.05
Let p(public) be the proportion of alumni of the public university who attended at least one class reunion
p(private) be the proportion of alumni of the private university who attended at least one class reunion
Hypotheses are:
: p(public) = p(private)
: p(public) ≠ p(private)
The formula for the test statistic is given as:
z=
where
- p1 is the sample proportion of public university students who attended at least one class reunion (
)
- p2 is the sample proportion of private university students who attended at least one class reunion (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the alumni from public university (1311)
- n2 is the sample size of the students from private university (1038)
Then z=
=-0.207
Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.
