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3 years ago

11

Can someone tell me if I’m correct

1 answer:

coldgirl [10]3 years ago

4 0

Answer:

No, you are only partially correct. Select sqrt(12)/3 and sqrt(8).

Step-by-step explanation:

An irrational number is simply a number that results in a non-repeating and non-terminating decimal. A rational number is simply any number that can be expressed as a ratio of whole numbers. Using these definitions, let's go through each answer choice:

<u>- 1 / 3 --> Rational number</u> since ratio of -1 to 3

<u>0.6666 (repeating) --> Rational number</u> since common ratio results in repeating non terminating number. In this case, 2/3.

<u>sqrt(12) / 3 --> Irrational number</u> since we have the ratio (2/3) times sqrt(3). The sqrt(3) is irrational itself since it cannot be expressed as a whole number ratio, and results in a non-repeating and non-terminating decimal. sqrt(12) reduces into sqrt(4 * 3) which reduces into 2sqrt(3).

<u>(sqrt(2))^2 --> Rational number</u> since the square root operation is negated by the square operation leaving the whole number 2.

<u>sqrt(8) --> Irrational number</u> since we have 2 * sqrt(2), which itself cannot be expressed as a whole number ratio, and results in a non-repeating and non-terminating decimal. sqrt(8) reduces to sqrt(4 * 2) which reduces to 2 * sqrt(2)

Hence, our two irrational numbers are sqrt(12)/3 and sqrt(8).

Cheers.

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a. $y=6(1.7472)^x$

b. $y=6e^{0.558t}$

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Step-by-step explanation:

a.-Given the first term at $t_0$ is 6 and the second term at $t_3$ is 32.

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$y=ab^x$

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$y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472$

#Replace b in the population function:

$y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x$

Hence, the regression for the rabbit population as a function of time x is $y=6(1.7472)^x$

b. The exponential function in terms of base $e$ is usually expressed as:

$A=A_0e^{kt}$

Where:

$A_0$ -is the initial population at $t_o$

$A$ -is the population at time t.

$k-$ is the exponential growth constant.

$e-$ the exponent

Our function in terms of base exponent is rewritten as:

$y=A_0e^{kt}$

#Substitute with actual figures to solve for t:

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Hence, the regression equation in terms of base e is $y=6e^{0.558t}$

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-We know that $y=6e^{0.558t}.$

Therefore we calculate t as(take y=10001):

$y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951$

Hence, it takes approximately 13.3 months for the population to exceed 10000

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4 years ago

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Answer:

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Step-by-step explanation:

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