When the variable is on both sides of the equation it is perfectly acceptable to solve each side by itself

A marine aquarium has a small tank and a large tank, each containing only red and blue fish. In each tank, the ratio of red fish

Anna007 [38]

Answer:

Ratio of blue fish in the small tank to the red fish in large tank is 10 : 6279

Step-by-step explanation:

Let the number of red fish and blue fish in the large tank are x and y respectively.

Similarly ratio of red fish and blue fish in the small tank are x' and y' respectively.

Since in each tank ratio of the red fish to blue fish is 333 : 444

That means x : y = 333 : 444

Or $\frac{x}{y}=\frac{333}{444}$

⇒ $\frac{x}{y}=\frac{3}{4}$

⇒ y = $\frac{4x}{3}$ --------(1)

Similarly x' : y' = 333 : 444

⇒ $\frac{x'}{y'}=\frac{333}{444}$

⇒ $\frac{x'}{y'}=\frac{3}{4}$

⇒ x' = $\frac{3y'}{4}$ ------(2)

Ratio of the fish in large tank to the fish in small tank is 464646 : 555

So (x + y) : (x' + y') = 464646 : 555

$\frac{x+y}{x'+y'}=\frac{464646}{555}$

Now we replace the values of x and y' from equation (1) and equation (2)

$\frac{(x+\frac{4x}{3})}{(\frac{3y'}{4}+y')}=\frac{464646}{555}$

$\frac{\frac{7x}{3}}{\frac{7y'}{4}}=\frac{464646}{555}$

$\frac{4}{3}\times \frac{x}{y'}=\frac{464646}{555}$

$\frac{x}{y'}=\frac{3}{4}\times \frac{464646}{555}$

$\frac{x}{y'}=\frac{232323}{370}$

$\frac{y'}{x}=\frac{370}{232323}$

$\frac{y'}{x}=\frac{10}{6279}$

Therefore, ratio of blue fish in the small tank to the red fish in large tank is 10 : 6279

4 0

3 years ago

Paul Havlik promised his grandson Jamie that he would give him $7,100 7 years from today for graduating from high school. Assume

Dominik [7]

Answer:

$\large \boxed{\$4100.07}$

Step-by-step explanation:

The formula for the future value (FV) of an investment earning compound interest is

$FV = PV \left (1 + \frac{r}{n} \right )^{nt}$

where

PV = the present value (PV) of the money invested

r = the annual interest rate expressed as a decimal fraction

t = the time in years

n = the number of compounding periods per year

Data:

FV = $7100

r = 8 % = 0.08

t = 7 yr

n = 2

Calculation:

$\begin{array}{rcl}\\7100& =& PV \left (1 + \dfrac{0.08}{2} \right )^{2 \times 7}\\\\& =& PV (1 + 0.04)^{14}\\\\& =&PV (1.04)^{14}\\& =& PV(1.731676)\\PV& =& \dfrac{7100}{1.731676}\\\\& =& \mathbf{4100.07}\\\end{array}\\\text{The present value of the money is $\large \boxed{\mathbf{\$4100.07}}$}$

4 0

2 years ago

Find a negative angle that is co-terminal with the angle 158.3

jenyasd209 [6]

Answer:0.0

Step-by-step explanation:

4 0

2 years ago

Which of the following is a solution to 2sin^(2)x+sinx-1=0

lukranit [14]

Hello,

$2*sin^2(x)+sin(x)-1=0\\\Delta=1+4*2*1=9=3^2\\sin^2(x)=\frac{-1-3}{4} =-1\ (impossible)\\or\\sin^2(x)=\frac{-1+3}{4} =\dfrac{1}{2}\\sin(x)=\dfrac{\sqrt{2} }{2} \Longrightarrow (x=45^o\ or\ x=135^o)\\or\\sin(x)=-\dfrac{\sqrt{2} }{2} \Longrightarrow (x=315^o\ or\ x=225^o)\\$

Answer A : 45°

6 0

2 years ago

A homeowner finds that there is a 0.15 probability that a flashlight does not work when turned on. If she has three flashlights,

prohojiy [21]

Answer:

A) 0.386

B).0.325

Step-by-step explanation:

probability p will not work = 0.15

Q probability w5ill work = 0.85

Number of flashlight = 3

Probabilty of at least one=

1- probability of none

But probability of none= 3C0(p)^0(q)^3

= 1*(0.15)^0(0.85)^3

= 1*1*(0.614)

= 0.614

Probability of at least one=1-0.614

= 0.386

Probability of two flashlight work

= 3C1(0.15)^1(0.85)^2

= 3*(0.15)*(0.7225)

= 0.325

6 0

2 years ago