A rancher’s herd of 250 sheep grazes over a 40-acre pasture. He would like to find out how many sheep are grazing on each acre of the pasture at any given time, so he has some images of the pasture taken by the state department of agriculture’s aerial photography division. Here are three samples of the images. Sample 1. | 4
Sample 2. | 1
Sample 3. | 9
1.] What can the rancher conclude from these samples about how many sheep graze on each acre of the 40-acre pasture?
2.] If the sheep were equally "spread out" across all of the 40 acres, how many sheep would you expect to find on average on each acre?
3.] How do the sample statistics compare to the population mean and standard deviation?
4.] What margin of error is reasonable for these samples?

it would more likely be bald eagle because its has more in the box so about 75%.

step-by-step explanation:

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages

i think the answer would be c