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3 years ago

The number of patients who reported that a new drug had relieved their pain. discrete or continuous?

Mathematics

1 answer:

solong [7]3 years ago

4 0

Answer:

Continuously

Step-by-step explanation:

Continuously

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\dfrac{23}{50} 50 23 start fraction, 23, divided by, 50, end fraction of Xander's friends prefer to stay home on Friday nights

cricket20 [7]

Answer:

46%

Step-by-step explanation:

Given that:

23/ 50 of Xander's friend prefer to stay at home;

Total number of friend's = 50

Number who prefer to stay home = 23

Hence, percentage who prefer to stay at home :

(23 /50) * 100%

0.46 * 100%

= 46%

4 0

3 years ago

Find all the solutions for the equation:

Contact [7]

$2y^2\,\mathrm dx-(x+y)^2\,\mathrm dy=0$

Divide both sides by $x^2\,\mathrm dx$ to get

$2\left(\dfrac yx\right)^2-\left(1+\dfrac yx\right)^2\dfrac{\mathrm dy}{\mathrm dx}=0$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

$x\dfrac{\mathrm dv}{\mathrm dx}+v=\dfrac{2v^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=\dfrac{2v^2-v(1+v)^2}{(1+v)^2}$

$x\dfrac{\mathrm dv}{\mathrm dx}=-\dfrac{v(1+v^2)}{(1+v)^2}$

The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

$\implies v^2+2v+1=(a+b)v^2+cv+a$

$\implies a=1,b=0,c=2$

Then

$\displaystyle\int\frac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=\int\left(\frac1v+\frac2{v^2+1}\right)\,\mathrm dv=\ln|v|+2\tan^{-1}v$

On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

$\ln\left|\frac{y(x)}x\right|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

7 0

3 years ago

Given a mean of 52.3 and a standard deviation of 1.8, what is the z-score of the value 51.8 rounded to the nearest tenth?

postnew [5]

The z-score of a certain data point in the given statistical data is calculated through the equation,
z-score = (x - m) / sd
where x is the data point, m is the mean and sd is the standard deviation.
Substituting the known values,
z-score = (51.8 - 52.3) / 1.8
z-score = -0.277
Rounding the answer to the nearest tenth will give an answer of -0.3.

8 0

3 years ago

4x+23=9x-38 21y-1 value of y

PSYCHO15rus [73]

Answer:

4⅗.78 99⁹7./⅜87.5⅞8 /7 ⁰

5 0

3 years ago

You guys help me please I’m stuck

DanielleElmas [232]

The answer should be $40

6 0

3 years ago