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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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HIJK is a parallelogram because the midpoint of both diagonals is ____ which means the diagonals bisect each other.
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EXPLANATION

We can use either diagonals to determine the midpoint.

We use the midpoint formula

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( \frac{ - 2 + 4}{2} , \frac{ - 3+ 3}{2} )

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Suppose that 4% of the 2 million high school students who take the SAT each year receive special accommodations because of docum
dangina [55]

Answer:

a. 0.0122

b. 0.294

c. 0.2818

d. 30.671%

e. 2.01 hours

Step-by-step explanation:

Given

Let X represents the number of students that receive special accommodation

P(X) = 4%

P(X) = 0.04

Let S = Sample Size = 30

Let Y be a selected numbers of Sample Size

Y ≈ Bin (30,0.04)

a. The probability that 1 candidate received special accommodation

P(Y = 1) = (30,1)

= (0.04)¹ * (1 - 0.04)^(30 - 1)

= 0.04 * 0.96^29

= 0.012244068467946074580191760542164986632531806368667873050624

P(Y=1) = 0.0122 --- Approximated

b. The probability that at least 1 received a special accommodation is given by:

This means P(Y≥1)

But P(Y=0) + P(Y≥1) = 1

P(Y≥1) = 1 - P(Y=0)

Calculating P(Y=0)

P(Y=0) = (0.04)° * (1 - 0.04)^(30 - 0)

= 1 * 0.96^36

= 0.293857643230705789924602253011959679180763352848028953214976

= 0.294 --- Approximated

c.

The probability that at least 2 received a special accommodation is given by:

P (Y≥2) = 1 -P(Y=0) - P(Y=1)

= 0.294 - 0.0122

= 0.2818

d. The probability that the number among the 15 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

First, we calculate the standard deviation

SD = √npq

n = 15

p = 0.04

q = 1 - 0.04 = 0.96

SD = √(15 * 0.04 * 0.96)

SD = 0.758946638440411

SD = 0.759

Mean =np = 15 * 0.04 = 0.6

The interval that is two standard deviations away from .6 is [0, 2.55] which means that we want the probability that either 0, 1 , or 2 students among the 20 students received a special accommodation.

P(Y≤2)

P(0) + P(1) + P(2)

=.

P(0) + P(1) = 0.0122 + 0.294

Calculating P(2)

P(2) = (0.04)² * (1 - 0.04)^(30 - 2(

P(2) = 0.00051

So,

P(0) +P(1) + P(2). = 0.0122 + 0.294 + 0.00051

= 0.30671

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Step-by-step explanation:

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We are required to find an Integer that can be inserted on the blank line (I have used x) in the list.

An integer is defined as a positive or negative whole number.

Out of the given options, only -1 and -2 are integers. However:

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Therefore, it cannot be our result.

The correct answer is -2.

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