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3 years ago

PLEASE HELP!! A fair coin is flipped 3 times. What is the probability of the coin landing on heads 3 times? Why is that?

Mathematics

2 answers:

Leokris [45]3 years ago

6 0

Answer:

If the coin is fair, then the odds of getting heads or tails should be equal, 12. Then 3 tosses of tails will have a chance of 12⋅12⋅12=18.

sdas [7]3 years ago

3 0

Answer:

1/8 or 12.5%

Step-by-step explanation:

The odds of getting heads is 50% (1/2) each time.

So putting the odds to the power of three will provide the odds of getting heads 3 times.

1/2 * 1/2 * 1/2 = 1/8 or 12.5%

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What is the area of this triangle ?!

fomenos

Answer:

35.

Step-by-step explanation:

To find the area, multiply the length and the width.

A = L x W

7 x 5 = 35

6 0

2 years ago

Read 2 more answers

The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine

salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

$f(x)=\frac{e^{\frac{-x}{1000} }}{1000}$

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

b. P(1000<x<2000) = 0.2325

c. P(x<1000) = 0.6321

d. 105.4 hours

Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = $\int\limits^b_a {P(x)} \, dx$

Then, for the electronic component, probability will be:

P(a<x<b) = $\int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx$

P(a<x<b) = $\frac{1000}{1000}.e^{\frac{-x}{1000} }$

P(a<x<b) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

a. For a component to last more than 3000 hours:

P(3000<x<∞) = $e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}$

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = $e^{-3}$

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = $e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}$

P(1000<x<2000) = $e^{-2}-e^{-1}$

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = $e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}$

P(0<x<1000) = $e^{-1}-1$

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = $e^{\frac{-b}{1000} }-e^\frac{-a}{1000}$

0.1 = $1-e^\frac{-x}{1000}$

$-e^{\frac{-x}{1000} }=-0.9$

${\frac{-x}{1000} }=ln0.9$

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

5 0

3 years ago

Which is a correct two column proof. Given <4 and

vladimir1956 [14]

The initial statement is: QS = SU (1)

QR = TU (2)

We have to probe that: RS = ST

Take the expression (1): QS = SU

We multiply both sides by R (QS)R = (SU)R

But (QS)R = S(QR) Then: S(QR) = (SU)R (3)

From the expression (2): QR = TU. Then, substituting it in to expression (3):

S(TU) = (SU)R (4)

But S(TU) = (ST)U and (SU)R = (RS)U

Then, the expression (4) can be re-written as:

(ST)U = (RS)U

Eliminating U from both sides you have: (ST) = (RS) The proof is done.

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Answer:

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Answer:

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Step-by-step explanation:

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