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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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we have

3x^{2} -8x+5

Equate the expression to zero to find the roots

3x^{2} -8x+5=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

3x^{2} -8x=-5

Factor the leading coefficient

3(x^{2} -(8x/3))=-5

Complete the square. Remember to balance the equation by adding the same constants to each side.

3(x^{2} -(8x/3)+(16/9))=-5+(16/3)

3(x^{2} -(8x/3)+(16/9))=(1/3)

Rewrite as perfect squares

3(x-(4x/3))^{2}=(1/3)

(x-(4x/3))^{2}=(1/9)

square roots both sides

(x-\frac{4}{3})=(+/-) \sqrt{\frac{1}{9}}

x=\frac{4}{3}(+/-) \frac{1}{3}

the roots are

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

x=\frac{4}{3}- \frac{1}{3}=\frac{3}{3}=1

so

3x^{2} -8x+5=3(x-\frac{5}{3})(x-1)=(3x-5)(x-1)

therefore

<u>the answer is the option</u>

(3x-5)(x-1)

6 0
2 years ago
Read 2 more answers
A closed cylindrical can of fixed volume V has radius r.a) Find the surface area, S, as a function of r.b) What happens to the v
andrey2020 [161]

Answer:

Step-by-step explanation:

This question is incomplete; here is the complete question.

A closed cylindrical can of fixed volume V has radius r. (a) Find the surface area, S, as a function of r. (b) What happens to the value of S approaches to infinity? (c) Sketch a graph of S against r, if  V=10 cm³.

A closed cylindrical can of volume V is having radius r and height h.

a). Surface area of a cylinder is given by

S = 2(Area of the circular sides) + Lateral area of the can

S = 2πr² + 2πrh

S = 2πr(r + h)

b). Since surface area is directly proportional to radius of the can

S ∝ r

Therefore, when r approaches to infinity (r → ∞)

c). If V = 10 cm³ Then we have to graph S against r.

From the formula V = πr²h

10 = πr²h

h = \frac{10}{\pi r^{2}}

By placing the value of h in the formula of surface area,

S = 2\pi r(r+\frac{10}{\pi r^{2}})

Now we can get the points to plot the graph,

r       -2             -1         0       1            2

S    -13.72     -13.72     0    26.28    35.13

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3 years ago
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zhannawk [14.2K]

Answer:

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Step-by-step explanation:

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Answer:

Sales tax is $4.20

Step-by-step explanation:

70*6/100=4.2

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