All categories

3 years ago

If you flip a coin 90 times, how many times would you expect to land on tails?

Mathematics

1 answer:

GarryVolchara [31]3 years ago

5 0

Answer:

It would be expected to land on tails 45 times.

Step-by-step explanation:

If there is a 50% chance to land on tails, about half of the time it would land on heads.

You might be interested in

Create a real-life representation of the following multiplication problem. Explain what the result represents in the problem.

olya-2409 [2.1K]

The real-life problem will be:- The total number of chocolates is 24 and there are 3 boxes what will be the number of chocolates filled in each box. The number of chocolates in each box will be 8.

<h3>What is an expression?</h3>

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

The given expression is:-

$24\times \dfrac{1}{3}$ = 8

The real-life problem will be given the as:-The total number of chocolates is 24 and there are 3 boxes what will be the number of chocolates filled in each box. The number of chocolates in each box will be 8.

Hence the number of chocolates in each box will be 8.

To know more about Expression follow

brainly.com/question/723406

#SPJ1

7 0

2 years ago

James and Sarah went out to lunch. The price of lunch for both of them was $40. They tipped their server

irina1246 [14]

Answer:

20%

Step-by-step explanation:

3 0

3 years ago

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

Novosadov [1.4K]

Answer:

$A(t)=240-220e^{-\frac{t}{40}}$

Step-by-step explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

Volume of the tank = 240 liters
Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

$R_{in}=$ (concentration of salt in inflow)(input rate of fluid)

$R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}$

$R_{out}=$ (concentration of salt in outflow)(output rate of fluid)

$R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}$

Rate of change of the amount of salt in the tank:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{40}$

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

$\dfrac{dA}{dt}+\dfrac{A}{40}=6\\$The integrating factor: e^{\int \frac{1}{40}dt} =e^{\frac{t}{40}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{40}}+\dfrac{A}{40}e^{\frac{t}{40}}=6e^{\frac{t}{40}}\\(Ae^{\frac{t}{40}})'=6e^{\frac{t}{40}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{40}})'=\int 6e^{\frac{t}{40}} dt\\Ae^{\frac{t}{40}}=6*40e^{\frac{t}{40}}+C, $(C a constant of integration)\\Ae^{\frac{t}{40}}=240e^{\frac{t}{40}}+C\\$Divide all through by e^{\frac{t}{40}}\\A(t)=240+Ce^{-\frac{t}{40}}$