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3 years ago

13

Due to lace of funding,the number of students enrolled at the city college went from 7000 last year to 4000 this year . Fine the

percent decrease in enrollment

1 answer:

AlladinOne [14]3 years ago

7 0

There is 42.8% of decrease in enrollment comparing to previous year.

42.8% of 7000 is 3000 which is the difference. Just divide 3000/7000 to get the numbers

In other words, new enrollmeny is 57.1% of the previous one

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What is an integer between the square root of 30 and 4*Pi divided by 3

Andru [333]

$\sqrt{30}\approx5.5\\
\frac{4\pi}{3}\approx4.2$

It's 5.

3 0

3 years ago

Simon has more money than Kande. if Simon gave Kande K20, they would have the same amount. While if Kande gave Simon $22, Simon

motikmotik

Given parameters;

Let us solve this problem step by step;

Let us represent Simon's money by S

Kande's money by K

Simon has more money than Kande

S > K

if Simon gave Kande K20, they would have the same amount;

if Simon gives $20, his money will be S - 20 lesser;

When Kande receives $20, his money will increase to K + 20

S - 20 = K + 20 ------ (i)

While if Kande gave Simon $22, Simon would then have twice as much as Kande;

if Kande gave Simon $22, his money will be K - 22

Simon's money, S + 22;

S + 22 = 2(K - 22) ------ (ii)

Now we have set up two equations, let us solve;

S - 20 = K + 20 ---- i

S + 22 = 2(K - 22) ; S + 22 = 2K - 44 ---- ii

So, S - 20 = K + 20

S + 22 = 2K - 44

subtract both equations;

-20 - 22 = (k -2k) + 64

-42 = -k + 64

k = 106

Using equation i, let us find S;

S - 20 = K + 20

S - 20 = 106 + 20

S = 106 + 20 + 20 = 146

Therefore, Kande has $106 and Simon has $146

3 0

3 years ago

A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int

Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}$

By expanding the previous equation:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)$

The tank capacity and capacity rate of change given in gallons and gallons per minute are, respectivelly:

$V_{tank} = 220\\\frac{dV_{tank}(t)}{dt} = 0$

Since there is no accumulation within the tank, expression is simplified to this:

$c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}$

By rearranging the expression, it is noticed the presence of a First-Order Non-Homogeneous Linear Ordinary Differential Equation:

$V_{tank} \cdot \frac{dc(t)}{dt} + f_{out} \cdot c(t) = c_0 \cdot f_{in}$ , where $c(0) = 0 \frac{pounds}{gallon}$ .

$\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}$

The solution of this equation is:

$c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})$

The salt concentration after 8 minutes is:

$c(8) = 0.166 \frac{pounds}{gallon}$

The instantaneous amount of salt in the tank is:

$m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds$

3 0

3 years ago

Find the difference. Write your answer in simplest form as a fraction or mixed number.

Harrizon [31]

Answer:

-2

Step-by-step explanation:

4 0

3 years ago

Which ordered pair (p,r) is the solution to the given system of linear equations?

Misha Larkins [42]

Answer:

(p,r) = (1/3, 2/9)

Step-by-step explanation:

Here, we want to solve a system of equations

We can rewrite the second equation by dividing through by 2

So we have;

4p + 3r = 2

and

5p - 3r = 1

Add both equations:

9p = 3

p = 3/9

p = 1/3

Recall ;

5p - 3r = 1

3r = 5p - 1

Substitute the value or p here

3r = 5(1/3)-1

3r = 5/3 - 1

3r = 2/3

r = 2/9

So we have the solution set as;

(p,r) = (1/3 , 2/9)

8 0

3 years ago