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2 years ago

A fair coin is flipped two times. Which diagram shows all of the possible outcomes?

Mathematics

1 answer:

kirza4 [7]2 years ago

6 0

Answer:

Step-by-step explanation:

It is the only one that has both heads and tails equally on all options, instead of heads heads or tails tails. There aren't any fair coins that are like that

hope this helps :)

You might be interested in

What is 32 = 3.2c so what is the letter c

Elena-2011 [213]

Whatever you do to one side, you have to do to the other.
32=3.2c
We need c by itself, so we need to divide both sides by 3.2.
32/3.2=10
C=10
Let’s make sure c equals ten by putting in ten for c.
3.2(10)=32
32=32
C=10

4 0

3 years ago

Read 2 more answers

-2 4/5 + 1/4 equals what ?

vlada-n [284]

Hey there! :)

Since 2 4/5 is greater than 1/4, the answer is negative.

You can just subtract 2 4/5 and 1/4 and add the (-) sign

2 4/5 - 1/4

Change the fractions so they can have a common denominator: 20

2 4/5 = 2 16/20

1/4 = 5/20

2 16/20 - 5/20 = 2 11/20

Add the (-) sign → -2 11/20

Answer ⇒ -2 11/20

:)

5 0

3 years ago

7 measured data points have a sample mean of 1403 and a standard deviation of 27. Determine the best estimate of the mean value

mina [271]

Answer:

Step-by-step explanation:

Hello!

You have sample of n=7 with mean X[bar]= 1403 and standard deviation S=27 and are required to estimate the mean with a 95%CI.

Asuming this sample comes from a normal population I'll use a stuent t to estimate the interval (a sample of 7 units is too small for the standard normal to be accurate for the estimation):

[X[bar]± $t_{n-1;1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$ ]

$t_{n-1;1-\alpha /2} = t_{6;0.975}= 2.365$

[1403±2.365* $\frac{27}{\sqrt{7} }$ ]

[1378.87;1427.13]

The margin of error is the semiamplitude of the interval and you can calculate it as:

$d= \frac{Upbond-Lowbond}{2}= \frac{1427.13-1378.87}{2} = 24.13$

With a confidence level of 95% you'd expect that the real value of the mean is contained by the interval [1378.87;1427.13], the best estimate of the mean value is expected to be ± 24.13 of 1403.

I hope it helps!

4 0

3 years ago

Lola had $235.00 in her savings account. Three months later, Lola had

mars1129 [50]

Answer: decreased by 42%

Step-by-step explanation:

5 0

3 years ago

USA Today reported that about 20% of all people in the United States are illiterate. Suppose you take eight people at random off

Law Incorporation [45]

Answer:

a) Figure and code attached

b) $E(X)=\mu = np = 8*0.2= 1.6$

$Var(X) =\sigma^2= np(1-p) = 8*0.2*(1-0.2) = 1.28$

$Sd(X)=\sigma= \sqrt{1.28}= 1.131$

c) $P(X\geq 7) =0.97$

And we can calculate this with the complement rule.

$P(X \geq 7) = 1-P(X$

So then we have:

$P(X \leq 6) = 0.03$

And we are interested on the valueof n who satisfy this expression.

And for this we can verify this with the following code:

"=BINOM.DIST(6,E54,0.2,TRUE)"

And as we can see on the second figure attached the value who satisfy the condition would be n = 60.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=8, p=0.2)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We can use the following R code to generate the histogram for this case:

> x <- seq(0,8,by = 1)

> y <- dbinom(x,8,0.2)

> plot(x,y,type = "h",main="Histogram")

And as we can see we got the result on the figure attached. And the distribution seems to be skewed to the right.

Part b

For this case the expected value is given by:

$E(X)=\mu = np = 8*0.2= 1.6$

The variance is given by:

$Var(X) =\sigma^2= np(1-p) = 8*0.2*(1-0.2) = 1.28$

And the standard deviation would be:

$Sd(X)=\sigma= \sqrt{1.28}= 1.131$

Part c

For this case we have the following inequality:

$P(X\geq 7) =0.97$

And we can calculate this with the complement rule.

$P(X \geq 7) = 1-P(X$

So then we have:

$P(X \leq 6) = 0.03$

And we are interested on the valueof n who satisfy this expression.

And for this we can verify this with the following code:

"=BINOM.DIST(6,E54,0.2,TRUE)"

And as we can see on the second figure attached the value who satisfy the condition would be n = 60.

5 0

3 years ago