Find the area of each triangle

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago

liz asked charlie to pick up her up a cup of frozen yogurt. she asked for I need to toppings from the six: chocolate chips, gran

Afina-wow [57]

Step-by-step explanation:

Total toppings = 6

Probability of chocolate chips = 1/6

Probability of raspberries = 1/6

And if both together then

Probability = 2/6 = 1/3

3 0

3 years ago

Assume that the population of human body temperatures has a mean of 98.6 degrees F and a standard deviation of 0.62 degrees F. I

dimulka [17.4K]

Answer:

0% probability of getting a mean temperature of 98.2 degrees F or lower.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 98.6, \sigma = 0.62, n = 106, s = \frac{0.62}{\sqrt{106}} = 0.06$

Find the probability of getting a mean temperature of 98.2 degrees F or lower.

This is the pvalue of Z when X = 98.2. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{98.2 - 98.6}{0.06}$

$Z = -6.67$

$Z = -6.67$ has a pvalue of 0.

So there is a 0% probability of getting a mean temperature of 98.2 degrees F or lower.

8 0

3 years ago

A cone-shaped paper cup can hold 35 cubic inches of food. If the radius of the cone is 2.4 inches, what is the height of the con

SVETLANKA909090 [29]

$\bf \textit{volume of a cone}\\\\
V=\cfrac{\pi r^2 h}{3}~~
\begin{cases}
r=radius\\
h=height\\
-----\\
r=2.4\\
V=35
\end{cases}\implies 35=\cfrac{\pi (2.4)^2 h}{3}\implies 105=5.76\pi h
\\\\\\
\cfrac{105}{5.76\pi }=h\implies 5.805467091295 \approx h\implies 5.81 \approx h$

7 0

3 years ago

Read 2 more answers

Find the area lying outside r=6cos(theta) and inside r=3+3cos(theta)

Sloan [31]

The area bounded by the 2 parabolas is A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b...

the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves. 
2cosθ = 1 => θ = ±π/3 

A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ 
= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ 
= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] .. 
.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)] 
= 3θ/2 +sin(2θ) - sin(θ) 

Area = A(π/3) - A(-π/3) 
= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3) 
= π.

7 0

4 years ago