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3 years ago

ASAP Please help me!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

wariber [46]3 years ago

7 0

Answer:

Step-by-step explanation:

I am so sorry, but I actually don’t know how to do that. What grade is this? Dang it this is really hard.

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Find the volume of the sphere.

Ksju [112]

Answer:

$\frac{1}{48} \pi$ units cubed

Step-by-step explanation:

The volume of a sphere is: $\frac{4}{3}\pi r^{3}$

We know the radius r = 1/4, so we just plug this into the equation:

(4/3)* $\pi$ *(1/4)^3 = (4/3)* $\pi$ *(1/64) = $\frac{1}{48} \pi$

So, the answer is $\frac{1}{48} \pi$ units cubed.

4 0

4 years ago

Read 2 more answers

PLEASE HELP IM BEING TIMED

Triss [41]

Answer:

I can’t see it good

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

I don’t know how to find the equation for this problem

Elis [28]

Try y=3/2x + 3/2

rise over run is 12/8 which simplifies to 3/2

and the y intercept is 1.5 or 3/2

6 0

3 years ago

Statistics??? Construct a 95% confidence interval for the population mean. Assume the population has a normal distribution. A sa

antiseptic1488 [7]

Given:
μ = $3120, population mean
σ = $677, population standard deviation.
The population is normally distributed.
n = 20, sample size.

At the 95% confidence level, the confidence interval is
$( \mu - 1.96\frac{\sigma}{ \sqrt{n} } ,\, \mu + 1.96\frac{\sigma}{ \sqrt{n} })$

1.96(σ/√n) = 1.96(677/√20) = 296.71
Th confidence interval for the mean is
($2823.29, $3416.71)

Answer: ($2823.29, $3416.71)

6 0

3 years ago

A 50-gal tank initially contains 10 gal of fresh water. At t = 0, a brine solution

scZoUnD [109]

$\huge \mathbb{SOLUTION:}$

$\begin{array}{l} \textsf{Let }A(t)\textsf{ be the function which gives the amount} \\ \textsf{of the salt dissolved in the liquid in the tank at} \\ \textsf{any time }t. \textsf{ We want to develop a differential} \\ \textsf{equation that, when solved, will give us an} \\ \textsf{expression for }A(t). \\ \\ \textsf{The basic principle determining the differential} \\ \textsf{equation is} \\ \\ \end{array}$

$\boxed{ \footnotesize \begin{array}{l} \qquad\quad \quad\Large{\dfrac{dA}{dt} = R_{in} - R_{out}} \\ \\ \textsf{where:} \\ \\ \begin{aligned} \bullet\: R_{in} &= \textsf{rate of the salt entering} \\ &= \left({\footnotesize \begin{array}{c}\textsf{Concentration of} \\\textsf{salt inflow}\end{array}}\right) \times \small(\textsf{Input of brine}) \\ \\ \bullet\: R_{out} &= \textsf{rate of the salt leaving} \\ &= \left({\footnotesize \begin{array}{c}\textsf{Concentration of} \\\textsf{salt outflow}\end{array}}\right) \times \small(\textsf{Output of brine}) \end{aligned} \end{array}} \\ \\$

$\begin{array}{l} \textsf{On the problem, the amount of salt in the tank,} \\ A(t), \textsf{changes overtime is given by the differential} \\ \textsf{equation} \\ \\ \footnotesize A'(t) = \left(\dfrac{4\ \textsf{gal}}{1\ \textsf{min}}\right)\!\!\left(\dfrac{1\ \textsf{lb}}{1\ \textsf{gal}}\right) - \left(\dfrac{2\ \textsf{gal}}{1\ \textsf{min}}\right)\!\!\left(\dfrac{A(t)\ \textsf{lb}}{10 + (4 - 2)t\ \textsf{gal}}\right) \\ \\ \textsf{There's no salt in the tank (fresh water) at the} \\ \textsf{start, so }A(0) = 0. \textsf{ The amount of solution in the} \\ \textsf{tank is given by }10 + (4 -2)t, \textsf{so the tank will} \\ \textsf{overflow once this expression is equal to the total} \\ \textsf{volume or capacity of the tank.} \\ \\ 10 + (4 - 2)t = 50 \\ \\ \textsf{Solving for }t,\textsf{ we get} \\ \\ \implies \boxed{t = 20\textsf{ mins}} \\ \\ A'(t) = 4 - \dfrac{2A(t)}{10 + 2t} \\ \\ A'(t) = 4 - \dfrac{1}{5 + t} A(t) \\ \\ A'(t) + \dfrac{1}{5 + t} A(t) = 4 \\ \\ \textsf{This is a linear ODE with integrating factor} \\ \mu (t) = e^{\int \frac{1}{5 + t}\ dt} = e^{\ln |5 + t|} = 5 + t \\ \\ \textsf{Multiplying this to the ODE, we get} \\ \\ (5 + t)A'(t) + A(t) = 4(5 + t) \\ \\ [(5 + t)A(t)]' = 20 + 4t \\ \\ (5 + t)A(t) = 20t + 2t^2 + C \\ \\ \textsf{Since }A(0) = 0, \textsf{ we get } C = 0. \\ \\ A(t) = \dfrac{2t^2 + 20t}{t + 5} \\ \\ A(t) = 2t + 10 - \dfrac{50}{t + 5} \\ \\ \textsf{So the function that gives the amount of salt at} \\ \textsf{any given time }t,\textsf{ is given by} \\ \\ \implies A(t) = 2t + 10 - \dfrac{50}{t + 5} \\ \\ \textsf{The amount of salt in the tank at the moment} \\ \textsf{of overflow or at }t = 20\textsf{ mins is equal to} \\ \\ A(20) = 2(20) + 10 - \dfrac{50}{20 + 5} \\ \\ \implies \boxed{A = 48\ \textsf{gallons}} \end{array}$

$\Large \mathbb{ANSWER:}$

$\qquad\red{\boxed{\begin{array}{l} \textsf{a. }20\textsf{ mins} \\ \\ \textsf{b. }48\textsf{ gallons}\end{array}}}$

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3 years ago