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2 years ago

Solve r = 4/3(p-q) for p.

Mathematics

1 answer:

alekssr [168]2 years ago

4 0

R =4/3 (p-q)
3 x r = 4 (p-q)
(3 x r)/4 = p - q

(3 x r)/4 - q = p

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What’s the fraction of 65 percent

Shtirlitz [24]

65% as a fraction is 65/100 which can also be reduced to 13/20

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2 years ago

Will give brainliest answer

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2 years ago

Please help me with geometry

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Answer:

A. If the side lengths are the same, then a triangle is not scalene.

Step-by-step explanation:

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2 years ago

According to a 2018 survey by Bankrate, 20% of adults in the United States save nothing for retirement (CNBC website). Suppose t

nalin [4]

Answer:

0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they save nothing for retirement, or they save something. The probability of an adult saving nothing for retirement is independent of any other adult. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of adults in the United States save nothing for retirement (CNBC website).

This means that $p = 0.2$

Suppose that sixteen adults in the United States are selected randomly.

This means that $n = 16$

What is the probability that three or less of the selected adults have saved nothing for retirement?

This is:

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{16,0}.(0.2)^{0}.(0.8)^{16} = 0.0281$

$P(X = 1) = C_{16,1}.(0.2)^{1}.(0.8)^{15} = 0.1126$

$P(X = 2) = C_{16,2}.(0.2)^{2}.(0.8)^{14} = 0.2111$

$P(X = 3) = C_{16,3}.(0.2)^{3}.(0.8)^{13} = 0.2463$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0281 + 0.1126 + 0.2111 + 0.2463 = 0.5981$

0.5981 = 59.81% probability that three or less of the selected adults have saved nothing for retirement

4 0

2 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

1 year ago