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stellarik [79]
2 years ago
13

On 1st September 2014 there were 5400 trees planted in a wood.

Mathematics
1 answer:
Aliun [14]2 years ago
3 0

Answer:

1. a = 5400

2. r = 0.96

3. Percentage decrement = 65.4%

Step-by-step explanation:

Given

N = ar^t

Solving (a): Write down the value of a

a implies the first term

And from the question, we understand that the initial number of trees is 5400.

Hence,

a = 5400

Solving (b): Show that r = 0.96

Using

N = ar^t

When a = 5400, t = 1 i.e. the first year and N = 5184

Substitute these values in the above expression

5184 = 5400 * r¹

5184 = 5400 * r

5184 = 5400r

Solve for r

r = 5184/5400

r = 0.96

Solving (c): Show that the trees has decreased by over 65% in 2040

First, we need to calculate number of years (t) in 2040

t = 2040 - 2014

t = 26

Substitute 26 for t, 5400 for a and 0.96 for r in N = ar^t to get the number of trees left

N = 5400 * 0.96^26

N = 1868.29658019

N = 1868 (approximated)

Next, we calculate the percentage change as thus:

%Change = (Final - Initial)/Initial * 100%

Where the initial number of trees =5400 and final = 1868

%Change = (1868 - 5400)/5400 * 100%

%Change = -3532/5400 * 100%

%Change = -3532%/54

%Change = -65.4%

The negative sign indicates a decrements or reduction.

Hence, percentage decrement = 65.4% and this is over 65%

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The article "Expectation Analysis of the Probability of Failure for Water Supply Pipes"† proposed using the Poisson distribution
oksian1 [2.3K]

Answer:

a. P(X ≤ 5) = 0.999

b. P(X > λ+λ) = P(X > 2) = 0.080

Step-by-step explanation:

We model this randome variable with a Poisson distribution, with parameter λ=1.

We have to calculate, using this distribution, P(X ≤ 5).

The probability of k pipeline failures can be calculated with the following equation:

P(k)=\lambda^{k} \cdot e^{-\lambda}/k!=1^{k} \cdot e^{-1}/k!=e^{-1}/k!

Then, we can calculate P(X ≤ 5) as:

P(X\leq5)=P(0)+P(1)+P(2)+P(4)+P(5)\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\P(3)=1^{3} \cdot e^{-1}/3!=1*0.3679/6=0.061\\\\P(4)=1^{4} \cdot e^{-1}/4!=1*0.3679/24=0.015\\\\P(5)=1^{5} \cdot e^{-1}/5!=1*0.3679/120=0.003\\\\\\P(X\leq5)=0.368+0.368+0.184+0.061+0.015+0.003=0.999

The standard deviation of the Poisson deistribution is equal to its parameter λ=1, so the probability that X exceeds its mean value by more than one standard deviation (X>1+1=2) can be calculated as:

P(X>2)=1-(P(0)+P(1)+P(2))\\\\\\P(0)=1^{0} \cdot e^{-1}/0!=1*0.3679/1=0.368\\\\P(1)=1^{1} \cdot e^{-1}/1!=1*0.3679/1=0.368\\\\P(2)=1^{2} \cdot e^{-1}/2!=1*0.3679/2=0.184\\\\\\P(X>2)=1-(0.368+0.368+0.184)=1-0.920=0.080

4 0
2 years ago
The speed of a sparrow is x km/h in still air. When the
Semmy [17]

Answer:

x = 5m/s

Step-by-step explanation:

Distance flying out = 12 km  (headwind)

Distance flying back = 12 km (tailwind)

total distance = 12 + 12 =24 km

wind speed = 1km/h

speed going out (with headwind) = (x - 1) km/h

speed coming back (with tailwind) = (x + 1) km/h

Time taken to go out = distance going out / speed going out

= 12 / (x-1)

Time taken to come back = distance coming back / speed coming back

= 12 / (x+1)

total time = time taken to go out + time taken to come back

5 =[ 12/(x-1) ] + [ 12/(x-1)]

expanding this, we will get

5x² - 24x - 5 = 0

solving quadratic equation, we will get

x = -1/5 (impossible because speed cannot be negative)

or

x = 5 (answer)

4 0
2 years ago
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.


4 0
2 years ago
Read 2 more answers
When you select a random person, the probability that this person will go to Ches
insens350 [35]

Answer: a= 0.0231

Step-by-step explanation:

n=8

p=0.75

q=1 - 0.75 = 0.25

(p=x)

p(5) = 8C5 (0.25)^5 (0.75)^8-5

p(5)= 0.0231

Or 2.31%

6 0
2 years ago
The table below shows the functions to represent the sale f(t), in thousands, of four companies in t years.
Tcecarenko [31]
Hi there

the companies showed an exponential growth in sale are

Company C with
f(t) = 1.26t

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f(t) = 1.09t

Because
1+r where r is the rate of growth

the rate of growth of Company c is 1+26%

The rate of growth of company D is
1+9%


So it's option d

Good luck!
4 0
2 years ago
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