Response:
Detailed explanation:
Hello!
Stratified sampling involves the categorization of the population into subgroups based on pre-established criteria for the study. These subgroups consist of homogeneous units concerning the relevant characteristics. In this instance, individuals in the groups will represent only one of the two potential opinions (support or not support) and not both.
The researcher determines the sample size desired, considering several factors such as finances, material availability, and accessibility to experimental subjects (for instance, if they are endangered species, larger sample sizes may not be feasible).
One might conduct proportionate stratified sampling by selecting a proportion of respondents who answered "yes" along with those who answered "no."
In this sampling method, taking a specific proportion from each subgroup allows for a more straightforward extrapolation of results to the overall populations. For example, if you needed a sample size of n = 20, each stratum would ideally contain half, meaning 10 from the “yes” group and 10 from the “no” group.
I hope this is helpful!
The diagrams for parts A and C are included here. For part B, we have circle O. We begin by drawing two radii OA and OC, connecting points A and C to create chord AC. The radius intersects chord AC at point B, bisecting AC into equal segments AB and BC. This gives us two triangles, ΔOBA and ΔOBC, where OA equals OC (since they're radii), OB equals OB (by the reflexive property), and AB is equal to BC (as stated in the question). By applying the SSS triangle congruence criterion, we conclude that ΔOBA is congruent to ΔOBC, allowing us to deduce that ∡OBA equals ∡OBC, both measuring 90°. Thus, OB is perpendicular to AC. Moving on to part D, we again work with circle O and draw the two radii OA and OC, joining points A and C to create chord AC. The radius intersects AC at point B, where AB is perpendicular to AC, meaning ∡B equals 90°. We then consider the right triangles ΔOBA and ΔOBC, and given OA equals OC (the radii), and OB equals OB (reflexive property), we conclude through the HL triangle congruence that ΔOBA is congruent to ΔOBC. Consequently, we find BA equal to BC, thus OB bisects AC.
Step-by-step explanation:
a) 7!
In absence of any restrictions, the answer is 7! as it represents the permutations of all animals.
b) 4! x 3!
Considering there are 6 cats and 5 dogs, the first and last slots must be occupied by cats to ensure alternate arrangements. The only options available then are based on the arrangement of the cats among themselves and the dogs among themselves, yielding 4! permutations for the cats and 3! for the dogs, thus leading to a total of 4! x 3! arrangements.
c) 3! x 5!
Here, the arrangement of the dogs among themselves can occur in 3! ways. Considering the dogs as a singular “object,” we can arrange this unit with the 4 cats, providing 5! total arrangements possible, leading to 3! · 5! arrangement possibilities.
d) 2 x 4! x 3!
In this scenario, both cats and dogs must be grouped together, allowing positions where all cats come before the dogs or vice versa. As there are two configurations, the resultant count is 2 multiplied by both arrangements, resulting in 2 x 4! x 3!
The response indicates that the test comprises 10 questions worth 3 points each and 14 questions worth 5 points. If there are any queries, feel free to ask!!! Thank you!
