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Dmitry [639]
3 years ago
9

Suppose a candidate for public office is favored by only 48% of the voters. if a sample survey randomly selects 2500 voters, the

percentage in the sample who favor the candidate can be thought of as a measurement from a normal curve with a mean of 48% and a standard deviation of 1%. based on this information, how often would such a survey show that 50% or more of the sample favored the candidate?
Mathematics
1 answer:
Drupady [299]3 years ago
7 0
Mean=0.48
standard deviation=0.01
thus using the z-score:
P(x>0.5) we shall have the following:
z=(0.5-0.48)/0.01=2
thus
P(x>0.5)
=1-P(x<0.5)
=1-P(z<2)
=1-0.9772
=0.0228
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bagirrra123 [75]
Hi!

We can set up a proportion then cross multiply to solve this.

\frac{45}{x}  \frac{18}{100}

45 x 100 = 4500
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The answer is 250

Hope this helps! :)
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3 years ago
Given that the slope of the graphed linear function is -3, which formula could represent the function?
stepan [7]

Answer: y= -3(x+4)

Step-by-step explanation:

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2 years ago
Vivian red 14 fewer pages than she was assigned to read she read 60 Pages write and solve an equation to find how many pages P V
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P = 74

Step-by-step explanation:

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5 0
3 years ago
Find general solutions of the differential equation. Primes denote derivatives with respect to x.
zepelin [54]

Answer:

\mathbf{3x^2y^3+2xy^4=C}

Step-by-step explanation:

From the differential equation given:

6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0

The equation above can be re-written as:

6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0

(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial  x}}} \ \ \text{at each point of R}

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

M(x,y) = 6xy^3 +2y^4\  and \ N(x,y) = 9x^2 y^2 +8xy^3

So;

\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3

       \dfrac{\partial N}{\partial y }

Let's Integrate \dfrac{\partial F}{\partial x}= M(x,y) with respect to x

Then;

F(x,y) = \int (6xy^3 +2y^4) \ dx

F(x,y) = 3x^2 y^3 +2xy^4 +g(y)

Now, we will have to differentiate the above equation with respect to y and set \dfrac{\partial F}{\partial x}= N(x,y); we have:

\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0  \\ \\ g(y) = C_1

Hence, F(x,y) = 3x^2y^3 +2xy^4 +g(y)  \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1

Finally; the general solution to the equation is:

\mathbf{3x^2y^3+2xy^4=C}

5 0
2 years ago
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