Answer:
I think it's C
Step-by-step explanation:
Alright, so we'd use the combinations with repetition formula, so we choose from 4 schools to distribute to and distribute 8 blackboards. It's then
( 8+4-1)!/8!(4-1)!=11!/(3!*8!)=165
For at least one blackboard, we first distribute 1 to each school and then have 4 blackboards left, getting (4+4-1)!/4!(4-1)!=7!/(4!*3!)=35
Answer:
B is the correct answer!
Step-by-step explanation:
It starts at 6 and ends at 11.
Hope this helps & please mark brainiest....
Answer:
a. We reject the null hypothesis at the significance level of 0.05
b. The p-value is zero for practical applications
c. (-0.0225, -0.0375)
Step-by-step explanation:
Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.
Then we have 
, 
, 
and 
, 
, 
. The pooled estimate is given by
a. We want to test 
vs 
(two-tailed alternative).
The test statistic is 
and the observed value is 
. T has a Student's t distribution with 20 + 25 - 2 = 43 df.
The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value 
falls inside RR, we reject the null hypothesis at the significance level of 0.05
b. The p-value for this test is given by 
0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.
c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)
, i.e.,
where 
is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So
, i.e.,
(-0.0225, -0.0375)
Answer:
0.3
Step-by-step explanation:
/ =division
So 9 divided by 30=0.3
