In how many different ways can a man divide 7 different gifts among his 3 children if the eldest is to receive 3 gifts and the o
1 answer:
The order in which gifts are received doesn't matter - if child X gets toy 1 then toy 2, it's the same as giving child X toy 2 then toy 1 - so we are counting combinations.
The eldest child receives 3 of the 7 gifts, so they have
possible choices of gifts.
The next child receives 2 of the remaining 4 gifts, so they have
choices.
The last child receives the remaining 2 gifts, and there is only
way to select the gifts for them.
By the multiplication using, the total number of ways of distributing 7 gifts among 3 children in the prescribed way is 35 • 6 • 1 = 210.
You might be interested in
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)