You flip a fair coin 10,000 times. Approximate the probability thatthe difference between the number of heads and the number of tails is at most 100.

Answers

Answer 1
Answer:

Answer:

E[X] = (np) (p+(1-p))^(n-1)

With n =10,000 and p=0.50 we get

E[X]= (10,000*0.50) (0.50+(1–0.50))^(10,000–1)

E[X] = 5,000(1)

So assuming the coin is fair (p=50%), then we can expect to get heads 5,000 times when the coin is tossed 10,000 times.

Step-by-step explanation:


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What are 567 times 3

Answers

Answer:

1,701

Step-by-step explanation:

the answer is 1,701 :D

Nicole just lit a new candle and then let it burn all the way down to nothing. Theinitial length of the candle was 12 inches and the candle burned at a rate of 1.5 inches
per hour. Make a table of values and then write an equation for L, in terms oft,
representing the length of the candle remaining unburned, in inches, t hours after the
candle was lit.

Answers

Answer: first one is 12 then second one is 10.5 then the third 9 then the fourth one is 7.5

Step-by-step explanation:

Given: 44 - 2(3x + 4) = -18Prove: x = 9

Answers

Substitute 9 as x in the given function:
44-2(3(9)+4)=-18
Now check and see if the statement it true and simplify.
44-2(27+4)=-18
44-2(31)=-18
44-62=-18
-18=-18
You can see this is true !!
Show my work to answer this question .

Simplify the expression:
3+ – 5(4+ – 3v)

Answers

Answer:

The answer is

15v - 17

Step-by-step explanation:

3+ – 5(4+ – 3v) can be written as

3 - 5( 4 - 3v)

Expand and simplify

That's

3 - 20 + 15v

15v - 17

Hope this helps you

At a large bank, account balances are normally distributed with a mean of $1,637.52 and a standard deviation of $623.16. What is the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650?

Answers

Answer:

P(\bar X >1650)=P(Z>(1650-1637.52)/((623.16)/(√(400)))=0.401)

And we can use the complement rule and we got:

P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the bank account balances of a population, and for this case we know the distribution for X is given by:

X \sim N(1637.52,623.16)  

Where \mu=1637.52 and \sigma=623.16

Since the distribution of X is normal then the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, (\sigma)/(√(n)))

And we can use the z score formula given by:

z = (\bar X -\mu)/((\sigma)/(√(n)))

And using this formula we got:

P(\bar X >1650)=P(Z>(1650-1637.52)/((623.16)/(√(400)))=0.401)

And we can use the complement rule and we got:

P(Z>0.401) =1-P(Z<0.401) = 1-0.656= 0.344

Answer: the probability is 0.49

Step-by-step explanation:

Since the account balances at the large bank are normally distributed.

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = account balances.

µ = mean account balance.

σ = standard deviation

From the information given,

µ = $1,637.52

σ = $623.16

We want to find the probability that a simple random sample of 400 accounts has a mean that exceeds $1,650. It is expressed as

P(x > 1650) = 1 - P(x ≤ 1650)

For x = 1650,

z = (1650 - 1637.52)/623.16 = 0.02

Looking at the normal distribution table, the probability corresponding to the z score is 0.51

P(x > 1650) = 1 - 0.51 = 0.49

If m = 1/2, n= 1/4 , then m²+ n²

Answers

In fraction 5/16

In decimal 0.3125