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Do you agree with the researchers method of predicting the population? Why or why not?

Do you agree with the researchers method of predicting the population? Why or why-example-1
User Edd Grant
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2 Answers

3 votes

Answer:WARNIG THESE ANSWER DO NOT BELONG TO ME JUST GIVING YOU ALL THE ASNWER AT ONCES HOPE THIS HELPS

Explanation:

1) If the arithmetic sequence is linear then yes, you can find the sequence with the common difference.

For example, if the common difference is add one now and you know that the fifth term is the number six, then you would know that the sequence is two, three, four, five, six.

to work out the sequence going backward from the fifth term, you just need to subtract the amount that was previously added. For example, if the sequence adds three every time you go to the next term, then when figuring out the previous term just subtract three from the current term. So to figure out the fourth term, subtract three from the fifth term.

2) Exponential functions are defined for all real numbers, and geometric sequences are defined only for positive integers. Another difference is that the base of a geometric sequence (the common ratio) can be negative, but the base of an exponential function must be positive.

3) Yes, I agree with the researcher's method. All the values of in the table correspond to the values of n by using their formula.

Do you agree with the researchers method of predicting the population? Why or why-example-1
User Brenlla
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4.6k points
7 votes

Yes, I agree with the researcher's method. All the values of
a_(n) in the table correspond to the values of n by using their formula.

Explanation:

Step 1:

If we substitute the values of n in the researcher's equation and we get the right values of
a_(n), we can agree with the researcher's method.

The researcher's formula is
a_(n) = 1000(2)^(n-1). Here
a_(n) is the wombat population and n is the number of years.

Step 2:

When n = 1,
a_(1) = 1000(2)^(1-1) = 1000(1) = 1,000,

when n = 2,
a_(2) = 1000(2)^(2-1) = 1000(2) = 2,000,

when n = 3,
a_(3) = 1000(2)^(3-1) = 1000(4) = 4,000,

when n = 4,
a_(4) = 1000(2)^(4-1) = 1000(8) = 8,000,

when n = 5,
a_(5) = 1000(2)^(5-1) = 1000(16) = 16,000.

As all the values correspond to the values on the table, I agree with the researcher's method.

User Rohit Pandey
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5.4k points