All categories

3 years ago

What is the p(3,1,2) on rolls of a die

1 answer:

3 0

Number 3 is 1/6 to role a 1 on the second roll is 1/36= 1/6 X 1/6
to get 2 on the third role is 1/216 so the probability is 1/216

You might be interested in

Solve the system of equation -5y + 4x=49 7y + 2x = -23

mina [271]

Answer:

Step-by-step explanation:

4 0

2 years ago

Find the holes<br> <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bx%2B3%7D%7B%28x%2B1%29%28x-2%29%7D" id="TexFormula1" t

olga55 [171]

Answer:

Step-by-step explanation:

The "holes" are the x-es where function doesn't exist.

one must not divide by 0, so given function doesn't exis if the denominator is equal 0

The product {(x+1)×(x-2)} is equal 0 if any of factors {(x+1),(x-2)} is 0, so

x + 1 = 0 or x - 2 = 0

x = - 1 or x = 2

8 0

2 years ago

Solve 3x^2 - 20x - 7 =0

grin007 [14]

This involvea factoring the polynomial into binomials and solving.
So make it (3x+1)(x-7)
X equals 7 and -1/3

5 0

3 years ago

Read 2 more answers

A bag of marbles had 14 blue marbles and the rest were white. For a game, 16 marbles were chosen from the bag. Of these 16 marbl

Musya8 [376]

Answer:

dfghjjjjjjjjjhhgg

Step-by-step explanation:

6 0

3 years ago

In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean

Nastasia [14]

Answer:

a) $\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

b) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma$ represent the population standard deviation

n=80 represent the sample size

Solution to the problem

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

For this case we can calculate the mean like this:

$\bar X =\frac{46+54}{2}=50$

And the margin of error is given by:

$ME= \frac{54-46}{2}= 4$

The confidence level is 0.96 and the significance level is $\alpha=1-0.96=0.04$ and the value of $\alpha/2 =0.02$ and the margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$

We can calculate the critical value and we got:

$z_{\alpha/2} = 2.05$

And if we solve for the deviation like this:

$\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}$

And replacing we got:

$\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45$

Part b

For this case is the sample size is doubled the margin of error would be:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828$

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor $\sqrt{2}$

5 0

2 years ago