A group incoming first-year students at Major University were surveyed in order to determine the factors that influenced their decisions to choose to attend Major U. The survey revealed the following information. 55 said "great football team" (F) 51 said "excellent party reputation" (P) 61 said "it was the only place that would accept me" (O) 9 said "great football team" but didn't say "excellent party reputation" and didn't say "it was the only place that would accept me" 26 said "it was the only place that would accept me" and "great football team" and "excellent party reputation" 31 said "excellent party reputation" and "great football team" 8 said only "excellent party reputation" 4 said none of the above reasons Required:a. How many said "it was the only place that would accept me" and "excellent party reputation"? b. How many said "great football team "or "it was the only place that would accept me"? c. How many said "great football team "but no "excellent party reputation"? d. How many were surveyed?

Step-by-step explanation:

This is a problem of SETS.

Start by listing out important data:

1. Total that said F = 55

2. Total that said P = 51

3. Total that said O = 61

4. F only = 9

5. F ∩ P ∩ O = 26 [NOTE: If you were to draw a Venn Diagram, 26 would be in the innermost circle because it comprises all three categories]

6. F ∩ P = 31

7. P only = 8

8. Students that said none of the 3 reasons = 4

QUESTIONS

1. How many said O and P? In other words, find the intersect of O and P. Find O ∩ P

2. How many said either F or O? [Answer to be gotten using a venn diagram] Find F ∪ P which translates to "F union P"

3. How many said F without saying P? [Answer to be gotten from the venn diagram as well]

4. How many students in total were surveyed? [HINT: Remember to include the 4 students that had none of the three options]

answer: b

step-by-step explanation:

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40