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lara31 [8.8K]
3 years ago
5

a fair die is cast four times. Calculate the probability of obtaining exactly two 6's round to the nearest tenth of a percent

Mathematics
1 answer:
Mrac [35]3 years ago
6 0

|\Omega|=6^4=1296\\|A|=1\cdot1\cdot5\cdot5\cdot\dfrac{4!}{2!2!}=25\cdot6=150\\\\P(A)=\dfrac{150}{1296}\approx11.6\%

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Find the inverse of square root f(n) = n – 3.
pogonyaev

Answer:

f'(n) = n + 3

Step-by-step explanation:

Given

f(n) = n - 3

Required

The inverse

f(n) = n - 3

Replace f(n) with y

y = n - 3

Swap positions of y and n

n = y - 3

Make y the subject

y = n + 3

Replace y with f'(n)

f'(n) = n + 3

4 0
3 years ago
A swimming pool with a volume of 30,000 liters originally contains water that is 0.01% chlorine (i.e. it contains 0.1 mL of chlo
SpyIntel [72]

Answer:

R_{in}=0.2\dfrac{mL}{min}

C(t)=\dfrac{A(t)}{30000}

R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}

A(t)=300+2700e^{-\dfrac{t}{1500}},$  A(0)=3000

Step-by-step explanation:

The volume of the swimming pool = 30,000 liters

(a) Amount of chlorine initially in the tank.

It originally contains water that is 0.01% chlorine.

0.01% of 30000=3000 mL of chlorine per liter

A(0)= 3000 mL of chlorine per liter

(b) Rate at which the chlorine is entering the pool.

City water containing 0.001%(0.01 mL of chlorine per liter) chlorine is pumped into the pool at a rate of 20 liters/min.

R_{in}=(concentration of chlorine in inflow)(input rate of the water)

=(0.01\dfrac{mL}{liter}) (20\dfrac{liter}{min})\\R_{in}=0.2\dfrac{mL}{min}

(c) Concentration of chlorine in the pool at time t

Volume of the pool =30,000 Liter

Concentration, C(t)= \dfrac{Amount}{Volume}\\C(t)=\dfrac{A(t)}{30000}

(d) Rate at which the chlorine is leaving the pool

R_{out}=(concentration of chlorine in outflow)(output rate of the water)

= (\dfrac{A(t)}{30000})(20\dfrac{liter}{min})\\R_{out}= \dfrac{A(t)}{1500} \dfrac{mL}{min}

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.2- \dfrac{A(t)}{1500}

We then solve the resulting differential equation by separation of variables.

\dfrac{dA}{dt}+\dfrac{A}{1500}=0.2\\$The integrating factor: e^{\int \frac{1}{1500}dt} =e^{\frac{t}{1500}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{1500}}+\dfrac{A}{1500}e^{\frac{t}{1500}}=0.2e^{\frac{t}{1500}}\\(Ae^{\frac{t}{1500}})'=0.2e^{\frac{t}{1500}}

Taking the integral of both sides

\int(Ae^{\frac{t}{1500}})'=\int 0.2e^{\frac{t}{1500}} dt\\Ae^{\frac{t}{1500}}=0.2*1500e^{\frac{t}{1500}}+C, $(C a constant of integration)\\Ae^{\frac{t}{1500}}=300e^{\frac{t}{1500}}+C\\$Divide all through by e^{\frac{t}{1500}}\\A(t)=300+Ce^{-\frac{t}{1500}}

Recall that when t=0, A(t)=3000 (our initial condition)

3000=300+Ce^{0}\\C=2700\\$Therefore:\\A(t)=300+2700e^{-\dfrac{t}{1500}}

3 0
3 years ago
ANSWER ONLY IF KNOW!! NO LINKS OR FAKE ANSWERS!!!<br> BRAINLIEST AND 30 POINTS!!!!!
koban [17]

Answer:

9 sq units

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What is the 308th term of the sequence 11,12.5,14,15.5,17...
Alecsey [184]

\huge\boxed{471.5}

The formula for an arithmetic sequence is as follows:

a_n=a_1+(n-1)d

In this formula, a_n represents the nth term of the sequence and d represents the common difference between each term.

Plug in the known values.

a_{308}=11+(308-1)1.5

Subtract.

a_{308}=11+307*1.5

Multiply.

a_{308}=11+460.5

Add.

a_{308}=\boxed{471.5}

6 0
2 years ago
TIME LIMIT PLEASE HELP ASAP
SashulF [63]

Answer:

y = 8, x = 4

Step-by-step explanation:

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