A survey of 300 parks showed the following. 15 had only camping. 20 had only hiking trails. 35 had only picnicking. 185 had camping. 140 had camping and hiking trails. 125 had camping and picnicking. 210 had hiking trails. Determine the number of parks that: a. Had at least one of these features.
b. Had all three features.
c. Did not have any of these features.
d. Had exactly two of these features.

a. Had at least one of these features = 290

b. Had all three features = 95

c. Did not have any of these features = 10

d. Had exactly two of these features = 125

Step-by-step explanation:

Given that;

Total parks = 300

15 had only camping

20 had only hiking trails

35 had only picnicking

-> 185 had camping

15 + a + d + c = 185 ------lets say equation 1

->  140 had camping and hiking trails

a + d = 140 ------------- lets say equation 2

now lets substitute equation 2 in 1

15 + 140 + c = 185

c = 30

->  125 had camping and picnicking

c + d = 125 lets say equation 3

substitute  c = 30 into equation 3

30 + d = 125

d = 95

Also substitute d = 95 in equation 2

a + 95 = 140

a = 45

->  210 had hiking trails

a  + b + d + 20 = 210

45 + b + 95 + 20 = 210

b = 50

Now e ( which is dont have any features)

= Total - (15 + 20 + 35 + a  + b + c + d)

= 300 - (15 + 20 + 35 + 30  + 50 + 45 + 95)

= 300 - 290

e = 10

so

a) Had at least one of these features :

Total - e

300 - 10 = 290

b) Had all three features :

Had all three features is d

d = 95

c) Did not have any of these features;

Did not have any of these features is e

e = 10

d) Had exactly two of these features;    

a + b + c

30 + 50 + 45

= 125

answer: the proof is mentioned below.

step-by-step explanation:

since, given:   lines bc = ef

prove:   lines ac > ef

          statements                                           reason

1. bc=ef                                           1. given

2. ac= ab+bc                                 2. because b is the point between the  

                                                            line segment ac

3.   ac > bc                                       3. because if a line segment is equal to

                                                          the sum of two line segments then it is

                                                          always greater then each of these

                                                          individual line segments

4. ac > ef                                         4. by equality property.