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Advocard [28]
3 years ago
10

Suppose you want to simulate surveying 100 people to find the date of the month (1st, 2nd, 3rd, . . .) each was born. You decide

to simulate the survey by having your computer randomly list values 1 through 31. Would the simulation provide a fair representation of an actual survey of 100 people? Why or why not?
Yes, because all possible dates of the month would be represented.
No, because for the 12 months in the year not all of the dates are represented. No, because for the 12 months in the year each date is not equally likely.
Yes, because it is possible to get 100 random values in a simulation.
Mathematics
1 answer:
ANTONII [103]3 years ago
4 0

Answer:

No, because for the 12 months in the year each date is not equally likely.

Step-by-step explanation:

Not all months have 31 days, meaning that numbers as high as 31 will not be as common.

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8 times 12 is 96 and 720 divided by 96 is 7.5. Your answer is 7.5
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WILL AWARD A MEDAL AND FIVE STARS!!!!
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Hello,

f(x)-2x-7

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3 years ago
A rectangle has a perimeter of 26 cm and one of its sides has length of 5 cm
otez555 [7]

Answer:5 by 8 by 5 by 8

Step-by-step explanation: the perimeter is 26 and one side if 5 so you know that another side is also 5. So add 5 and 5 together you get 10. Then subtract 10 from the perimeter(witch we know is 26) and you end up with 16. we know that the last two side lengths are the same so divide 16 by two and the sidle length is 8.

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3 years ago
Which definite integral approximation formula is this: the integral from a to b of f(x)dx ≈ (b-a)/n * [<img src="https://tex.z-d
Stella [2.4K]

The answer is most likely A.

The integration interval [<em>a</em>, <em>b</em>] is split up into <em>n</em> subintervals of equal length (so each subinterval has width (<em>b</em> - <em>a</em>)/<em>n</em>, same as the coefficient of the sum of <em>y</em> terms) and approximated by the area of <em>n</em> rectangles with base (<em>b</em> - <em>a</em>)/<em>n</em> and height <em>y</em>.

<em>n</em> subintervals require <em>n</em> + 1 points, with

<em>x</em>₀ = <em>a</em>

<em>x</em>₁ = <em>a</em> + (<em>b</em> - <em>a</em>)/<em>n</em>

<em>x</em>₂ = <em>a</em> + 2(<em>b</em> - <em>a</em>)/<em>n</em>

and so on up to the last point <em>x</em> = <em>b</em>. The right endpoints are <em>x</em>₁, <em>x</em>₂, … etc. and the height of each rectangle are the corresponding <em>y </em>'s at these endpoints. Then you get the formula as given in the photo.

• "Average rate of change" isn't really relevant here. The AROC of a function <em>G(x)</em> continuous* over an interval [<em>a</em>, <em>b</em>] is equal to the slope of the secant line through <em>x</em> = <em>a</em> and <em>x</em> = <em>b</em>, i.e. the value of the difference quotient

(<em>G(b)</em> - <em>G(a)</em> ) / (<em>b</em> - <em>a</em>)

If <em>G(x)</em> happens to be the antiderivative of a function <em>g(x)</em>, then this is the same as the average value of <em>g(x)</em> on the same interval,

g_{\rm ave}=\dfrac{G(b)-G(a)}{b-a}=\dfrac1{b-a}\displaystyle\int_a^b g(x)\,\mathrm dx

(* I'm actually not totally sure that continuity is necessary for the AROC to exist; I've asked this question before without getting a particularly satisfying answer.)

• "Trapezoidal rule" doesn't apply here. Split up [<em>a</em>, <em>b</em>] into <em>n</em> subintervals of equal width (<em>b</em> - <em>a</em>)/<em>n</em>. Over the first subinterval, the area of a trapezoid with "bases" <em>y</em>₀ and <em>y</em>₁ and "height" (<em>b</em> - <em>a</em>)/<em>n</em> is

(<em>y</em>₀ + <em>y</em>₁) (<em>b</em> - <em>a</em>)/<em>n</em>

but <em>y</em>₀ is clearly missing in the sum, and also the next term in the sum would be

(<em>y</em>₁ + <em>y</em>₂) (<em>b</em> - <em>a</em>)/<em>n</em>

the sum of these two areas would reduce to

(<em>b</em> - <em>a</em>)/<em>n</em> = (<em>y</em>₀ + <u>2</u> <em>y</em>₁ + <em>y</em>₂)

which would mean all the terms in-between would need to be doubled as well to get

\displaystyle\int_a^b f(x)\,\mathrm dx\approx\frac{b-a}n\left(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n\right)

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3 years ago
Which equation represents the circle described?
bazaltina [42]

We have the equation:

x^2+y^2-8x-6y+24=0


By arranging this equation in terms of x and y, we have:


x^2-8x+y^2-6y=-24 \\ \\


By using the method of completing the square, we have:

x^2-8x+\mathbf{\left(\frac{8}{2}\right)^2}+y^2-6y+\mathbf{\left(\frac{6}{2}\right)^2}=-24+\mathbf{\left(\frac{8}{2}\right)^2}+\mathbf{\left(\frac{6}{2}\right)^2} \\ \\ x^2-8x+\mathbf{16}+y^2-6y+\mathbf{9}=-24+\mathbf{16}+\mathbf{9} \\ \\ \boxed{(x-4)^2+(x-3)^2=1}


The center of this circle is:

(h,k)=(4,3)


So the equation that fulfills the statement is:

(x-4)^2+(y-3)^2=2^2


Finally, the right answer is c)

6 0
3 years ago
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