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Elza [17]
3 years ago
12

(3,-2), (-3,-2), (1,4), (-6,5), (1,3), (-20, 7)} Domain : Range:

Mathematics
1 answer:
dexar [7]3 years ago
6 0
Domain: -20, -6, -3, 1, 1, 3 ( in order from least to great )

range: -2, -2, 3, 4, 5, 7 ( in order from greatest to least )
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Hi! i need help<br> (It is due in a few hours)
wel

Answer:

A and A is the answer. so this is the answer

7 0
3 years ago
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What is the percent of 14536?
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1453.6000000000001 That is the percent
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3 years ago
The number and frequency of Atlantic hurricanes annually from 1940 through 2007 is shown here:
klio [65]

Answer:

The probability table is shown below.

A Poisson distribution can be used to approximate the model of the number of hurricanes each season.

Step-by-step explanation:

(a)

The formula to compute the probability of an event <em>E</em> is:

P(E)=\frac{Favorable\ no.\ of\ frequencies}{Total\ NO.\ of\ frequencies}

Use this formula to compute the probabilities of 0 - 8 hurricanes each season.

The table for the probabilities is shown below.

(b)

Compute the mean number of hurricanes per season as follows:

E(X)=\frac{\sum x f_{x}}{\sum f_{x}}=\frac{176}{68}=  2.5882\approx2.59

If the variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 7.56 then the probability function is:

P(X=x)=\frac{e^{-2.59}(2.59)^{x}}{x!} ;\ x=0, 1, 2,...

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-2.59}(2.59)^{0}}{0!} =\frac{0.075\times1}{1}=0.075

Compute the probability of <em>X</em> = 1 as follows:

\neq P(X=1)=\frac{e^{-2.59}(2.59)^{1}}{1!} =\frac{0.075\times7.56}{1}=0.1943

Compute the probabilities for the rest of the values of <em>X</em> in the similar way.

The probabilities are shown in the table.

On comparing the two probability tables, it can be seen that the Poisson distribution can be used to approximate the distribution of the number of hurricanes each season. This is because for every value of <em>X</em> the Poisson probability is approximately equal to the empirical probability.

5 0
3 years ago
1. The number of rabbits on an island is increasing exponentially.
Zina [86]

Answer:

<em>There are approximately 114 rabbits in the year 10</em>

Step-by-step explanation:

<u>Exponential Growth </u>

The natural growth of some magnitudes can be modeled by the equation:

P=P_o(1+r)^t

Where P is the actual amount of the magnitude, Po is its initial amount, r is the growth rate and t is the time.

We are given two measurements of the population of rabbits on an island.

In year 1, there are 50 rabbits. This is the point (1,50)

In year 5, there are 72 rabbits. This is the point (5,72)

Substituting in the general model, we have:

50=P_o(1+r)^1

50=P_o(1+r)\qquad\qquad[1]

72=P_o(1+r)^5\qquad\qquad[2]

Dividing [2] by [1]:

\displaystyle \frac{72}{50}=(1+r)^{5-1}=(1+r)^{4}

Solving for r:

\displaystyle r=\sqrt[4]{\frac{72}{50}}-1

Calculating:

r=0.095445

From [1], solve for Po:

\displaystyle P_o=\frac{72}{(1+r)^5}

\displaystyle P_o=\frac{72}{(1.095445)^5}

P_o=45.64355

The model can be written now as:

P=45.64355(1.095445)^t

In year t=10, the population of rabbits is:

P=45.64355(1.095445)^{10}

P = 113.6

P\approx 114

There are approximately 114 rabbits in the year 10

5 0
3 years ago
ASAP BRAINLIST . Answer question in the picture
MakcuM [25]

Answer:

this is non linear because it is not a straight line

Step-by-step explanation:

3 0
3 years ago
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