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3 years ago

What number is fifteen sixty percent of

1 answer:

6 0

I'm pretty sure it is 25. A method you could use(calculator would be more quicker, but by hand works too) is dividing 15 by sixty percent. To do that, u find the decimal of sixty percent. You would move the decimal point of sixty(the original place of the decimal would be 60.0) two places to the left. Then 60% would become 0.60. But, you could simplify that to 0.6, then you do 15 divided 0.6. You would get 25. If you are doing it by hand, double check the answer by multiplying it with the decimal and you should get the original number(which is 15 in this case). I hope this helped :)

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3 years ago

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In a recent survey, 60% of the community favored building a health center in their neighborhood. If 14 citizens are chosen, find

andreev551 [17]

Answer:

Probability that exactly 5 of them favor the building of the health center is 0.0408.

Step-by-step explanation:

We are given that in a recent survey, 60% of the community favored building a health center in their neighborhood.

Also, 14 citizens are chosen.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 14 citizens

r = number of success = exactly 5

p = probability of success which in our question is % of the community

favored building a health center in their neighborhood, i.e; 60%

<em>LET X = Number of citizens who favored building of the health center.</em>

So, it means X ~ $Binom(n=14, p=0.60)$

Now, Probability that exactly 5 of them favor the building of the health center is given by = P(X = 5)

P(X = 5) = $\binom{14}{5} \times 0.60^{5} \times (1-0.60)^{14-5}$

= $2002 \times 0.60^{5} \times 0.40^{9}$

= 0.0408

Therefore, Probability that exactly 5 of them favor the building of the health center is 0.0408.

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3 years ago

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One

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$y'=(t+y)^2-1$

Substitute $u=t+y$ , so that $u'=y'$ , and

$u'=u^2-1$

which is separable as

$\dfrac{u'}{u^2-1}=1$

Integrate both sides with respect to $t$ . For the integral on the left, first split into partial fractions:

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

$\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt$

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$ :

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$ :

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

Now use the given initial condition to solve for $C$ :

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$

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3 years ago