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lisov135 [29]
2 years ago
11

There are frogs and koi in a pond, and the number of frogs and the number of koi in the pond are independent. Let X represent th

e number of frogs in any given week, and let Y represent the number of koi in any given week. X has a mean of 28 with a standard deviation of 2.7, and Y has a mean of 15 with a standard deviation of 1.6. Which answer choice correctly calculates and interprets the standard deviation of the difference, D = X - Y?
Sigma Subscript D = 1.05; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean.
Sigma Subscript D = 1.1; this pond can expect the difference of frogs and koi to vary by approximately 1.1 from the mean.
Sigma Subscript D = 3.1; this pond can expect the difference of frogs and koi to vary by approximately 3.1 from the mean.
Sigma Subscript D = 13; this pond can expect the difference of frogs and koi to vary by approximately 1.05 from the mean.
Mathematics
1 answer:
brilliants [131]2 years ago
5 0

Answer: = 3.1; this pond can expect the difference of frogs and koi to vary by approximately 3.1 from the mean.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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For this case we have the following equation:

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We apply distributive property on the left side of the equation:

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We simplify the left side of the equation:

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Different signs are subtracted and the major sign is placed.

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On the right side we must take into account that:

- * + = -

So:

\frac {5} {3} x + 2 = 4-x-1\\\frac {5} {3} x + 2 = 3-x

We add x to both sides of the equation:

\frac {5} {3} x + x + 2 = 3\\\frac {5 * 1 + 3 * 1} {3} x + 2 = 3\\\frac {5 + 3} {3} x + 2 = 3\\\frac {8} {3} x + 2 = 3

We subtract 2 from both sides of the equation:

\frac {8} {3} x = 3-2\\\frac {8} {3} x = 1

We multiply by 3 on both sides of the equation:

8x = 3

We divide by 8 on both sides of the equation:

x = \frac {3} {8}

Thus, the solution of the equation is:

x = \frac {3} {8}

Answer:

x = \frac {3} {8}

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Step-by-step explanation:

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