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3 years ago

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1. What is the theoretical probability that a coin toss results in two heads showing?

1 answer:

Setler [38]3 years ago

8 0

Answer:

Step-by-step explanation:

1. What is the theoretical probability that a coin toss results in two heads showing?

25%

2. What is the experimental probability that a coin toss results in two heads showing?

50%

3. What is the theoretical probability that a coin toss results in two tails showing

0.5%

4. What is the experimental probability that a coin toss results in two tails showing?

25%

5. What is the theoretical probability that a coin toss results in one head and one tail showing?

0.5

6. What is the experimental probability that a coin toss results in one head and one tail showing?

0.5

7. Compare the theoretical probabilities to your experimental probabilities. Why might there be a difference?

0.5

There you go. Remember you can figure these answers out by yourself if you really try. From now on ill be answering questions and trying to help other people. Please read from your lessons as they can help you a lot, Thank you.

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What percent of 7 is 2?

Liono4ka [1.6K]

It is 350% I think, because you are trying to find how many groups of 2 go into 7. So, you have to divide 7 by 2. You get 3.5. 3.5 in percentage form is 350%

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3 years ago

What would the bottles purchased be and what would the bottles recycled be?

N76 [4]

Part A

Your problem statement already shows you the conversion.

Bottles purchased: 5×10¹⁰

Bottles recycled: 8.3×10⁹

Part B

The ratio of bottles purchased to bottles recycled is

... (5×10¹⁰)/(8.3×10⁹) = 0.6×10¹ = 6.0

The number of bottles purchased is 6 times the number of bottles recycle.

5 0

3 years ago

Adrian works in New York City and makes $40 per hour. She works in an office and must pay $20 a day for lunch. If she wants to m

andrew-mc [135]

Answer: She would have to work 7 hours to earn $260.

Step-by-step explanation: 40×7=280 280-20=260 Hope this helps.

6 0

3 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago

What is the proper equation <br> (-5cd^-4)(2cd^2)^2

mojhsa [17]

Answer:

-20c^3

Step-by-step explanation:

((−5c)(d−4))((2cd2)2)

= −20c3d4 /d4

=-20c^3

7 0

3 years ago