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lesya692 [45]
2 years ago
14

Tammy has planted a small, young tree in her yard. To allow for the tree’s growth, she needs 15 feet in all directions around th

e base of the tree to remain open and unplanted. Which best describes the section of ground surrounding the base of the tree that should remain unplanted?
Mathematics
1 answer:
astraxan [27]2 years ago
7 0

Answer:

First, remember that a circle of radius R centered in the point (a, b) can be written as:

(x - a)^2 + (y - b)^2  = R^2

Suppose that we can model the yard as a rectangular coordinate axis.

And the tree is planted in the point (a, b)

If we want to have 15 feet in all directions around the base of the tree (15 ft around the point (a, b))

The section that must remain unplanted is:

(x - a)^2 + (y - b)^2   ≤ R^2

Where the ≤ symbol is used because all the interior of the circle must remain unplanted (border included)

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Which inequality represent all values of x for which the quotient below is defined??
Liono4ka [1.6K]

Answer:

the answer is b. because x is either grater or equal.

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Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 pro
bearhunter [10]

The question is incomplete. Here is the complete question:

Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.95 probability that he will hit it. One day, Samir decides to attempt to hit  10 such targets in a row.

Assuming that Samir is equally likely to hit each of the 10 targets, what is the probability that he will miss at least one of them?

Answer:

40.13%

Step-by-step explanation:

Let 'A' be the event of not missing a target in 10 attempts.

Therefore, the complement of event 'A' is \overline A=\textrm{Missing a target at least once}

Now, Samir is equally likely to hit each of the 10 targets. Therefore, probability of hitting each target each time is same and equal to 0.95.

Now, P(A)=0.95^{10}=0.5987

We know that the sum of probability of an event and its complement is 1.

So, P(A)+P(\overline A)=1\\\\P(\overline A)=1-P(A)\\\\P(\overline A)=1-0.5987\\\\P(\overline A)=0.4013=40.13\%

Therefore, the probability of missing a target at least once in 10 attempts is 40.13%.

6 0
1 year ago
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A special 8-sided die is marked with the numbers 1 to 8. It is rolled 20 times with these outcomes:
horsena [70]

Answer:

One: B

Two: 60% and 10%

Step-by-step explanation:

Problem One

There are only two numbers in the sample of 40 that are under 26. Both are 25. If you find more, make the adjustment. There are 2 more that are exactly 26 but they are not counted because the directions say "less than 26."

So set up your proportion

x/2000 = 2/40                     Multiply both sides by 2000

x = 2/40 * 2000

x = 4000/40

x = 100

A

I don't know where 5 comes from. But it is not correct.

B

B should be the correct answer.

C

Exactly 100 pieces should be defective. That is the theoretical result. C is incorrect.

D

D is not correct. The sample size would not be 40. It would have to be 2000 for D to be correct. So D is wrong.

E

We have enough data to get an answer. E is incorrect.

Problem 2

The think you must NOT do is count 1 as being prime. The prime numbers are 2 3 5 7 between 1 and 8. They break down as follows.

  • Prime           Number of them
  • 2                         3
  • 3                         4
  • 5                         2
  • 7                         3

The total number of primes = 12

There are 20 numbers in the sample

The experimental probability of tossing a prime is 12/20 * 100% = 60%

The non primes are 2 3 5 7 which is 4 out of 8

4/8 * 100 = 50%

The experimental value is 10% more than the theoretical value.

Discussion

Note: the problem may be one. This all depends on what you have been told about 1. I am using the exact wording of prime here. 1 is not a prime. It is also not a composite. So it has to be counted as part of the non primes.


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Using this net speed, we can already calculate for the total time:

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