Choose what the expressions below best represent within the context of the word problem. The length of a rectangle is 12 inches
2 answers:
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Answer:
d. t distribution with df = 80
Step-by-step explanation:
Assuming this problem:
Consider independent simple random samples that are taken to test the difference between the means of two populations. The variances of the populations are unknown, but are assumed to be equal. The sample sizes of each population are n1 = 37 and n2 = 45. The appropriate distribution to use is the:
a. t distribution with df = 82.
b. t distribution with df = 81.
c. t distribution with df = 41.
d. t distribution with df = 80
Solution to the problem
When we have two independent samples from two normal distributions with equal variances we are assuming that
And the statistic is given by this formula:
Where t follows a t distribution with degrees of freedom and the pooled variance
is given by this formula:
This last one is an unbiased estimator of the common variance
So on this case the degrees of freedom are given by:
And the best answer is:
d. t distribution with df = 80
Answer:
a) 182 possible ways.
b) 5148 possible ways.
c) 1378 possible ways.
d) 2899 possible ways.
Step-by-step explanation:
The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.
Combinations formula:
is the number of different combinations of x objects from a set of n elements, given by the following formula.
In this question, we have that:
There are 52 total cards, of which:
13 are spades.
13 are diamonds.
13 are hearts.
13 are clubs.
(a)Two-pairs: Two pairs plus another card of a different value, for example:
2 pairs of 2 from sets os 13.
1 other card, from a set of 26(whichever two cards were not chosen above). So
So 182 possible ways.
(b)Flush: five cards of the same suit but different values, for example:
4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So
So 5148 possible ways.
(c)Full house: A three of a kind and a pair, for example:
4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).
3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So
So 1378 possible ways.
(d)Four of a kind: Four cards of the same value, for example:
4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).
1 from the remaining 39(do not involve the kind chosen above). So
So 2899 possible ways.