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3 years ago

Which percentage is the right answer

Mathematics

1 answer:

givi [52]3 years ago

3 0

This is a bad question.

The real answer is any of them may be the actual percentage, the true population mean. Probability and statistics often don't rule events out entirely; they just tell us some events are very unlikely. For example here all we know is there's a 95% chance the true percentage is within $78\% \pm 8\%$ . But there's a 5% chance it isn't, so those other percentages are still possible.

The question seems to want to ask which percentage falls within the 95% confidence interval. That's between 78-8=70% and 78+8=86% so the answer is

Answer: C 72%

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The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i

harina [27]

Answer:

$y(t) = y_o e^{\beta t}$

$x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

Step-by-step explanation:

From the question we are told that

$\frac{dy}{y} = -\beta dt$

Now integrating both sides

$ln y = \beta t + c$

Now taking the exponent of both sides

$y(t) = e^{\beta t + c}$

=> $y(t) = e^{\beta t} e^c$

Let $e^c = C$

$y(t) = C e^{\beta t}$

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

$y(t) = y_o e^{\beta t}$

From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

substituting for y

$\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })$

=> $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

$lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

$x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

=> $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

=> $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

$x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

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4 years ago

2. CAN SOMEONE PLEASE HELP ME? I'M NOT GOOD IN MATH.<br>Explain your work

Elodia [21]

Answer:

Step-by-step explanation:

Since triangles ABC and CDB are similar, CD and CB is the same as the ratio between CB and CA. 15:9=x:15. x=25. Subtracting CD to just be left with AD, we get 25-9=16, or answer choice D. Hope this helps!

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4 years ago

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Vaselesa [24]

The answer is c18.71

6 0

3 years ago

I need help, if you give answer ASAP I’ll reward a lot of points!!

Gnoma [55]

Answer:

A = 31.875 inches squared

Step-by-step explanation:

Area of a parallelogram is bh

b = 12 and 3/4 is just 12.75 inches

h = 2 and 1/2 is just 2.50 inches

Multiply these together and you get the area,

A = (12.75)(2.50) = 31.875 inches squared

4 0

3 years ago

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Can somebody help me please .

elena-14-01-66 [18.8K]

Get the question by itself

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3 years ago