Math work pls help :)

Suppose that two balanced, six sided dice are tossed repeatedly and the sum of the two uppermost faces is determined on each tos

Simora [160]

Answer:

$\frac{(2/36)}{(1-(28/36))} = 1/4$

Step-by-step explanation:

6 0

3 years ago

Jim types 1800 words in 25 minutes. How many words does he type per minute

kirill [66]

1800 divided by 25 is 72. He types 72 words per minute.

3 0

3 years ago

Answer this question PLEASEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

yarga [219]

Answer:

The last one.

Step-by-step explanation:

3 0

2 years ago

At one point the average price of regular unleaded gasoline was $3.41 per gallon. Assume that the standard deviation price per

Aleonysh [2.5K]

Answer:

a) $1-\frac{1}{k^2} =1- \frac{1}{2^2}= 1-0.25 = 0.75$

So we expected about 75% within two deviations from the mean

b) $1-\frac{1}{k^2} =1- \frac{1}{1.5^2}= 1-0.4444 = 0.556$

So we expected about 55.6% within 1.5 deviations from the mean

And the limits are:

$Lower = 3.41 -1.5*0.09 = 3.275$

$Upper = 3.41 +1.5*0.09 = 3.545$

c) We can calculate how many deviations we are within the mean with the limits with this formula:

$z =\frac{x-\mu}{\sigma}$

And using the lower limit we got:

$z = \frac{3.05-3.41}{0.09}=-4$

And with the upper limit we got:

$z = \frac{3.77-3.41}{0.09}=4$

So then the value of k =4 and the percentage is given by:

$1-\frac{1}{k^2} =1- \frac{1}{4^2}= 1-0.0625 = 0.9375$

Step-by-step explanation:

Previous concepts and Data given

$\mu =3.41$ reprsent the population mean

$\sigma=0.09$ represent the population standard deviation

The Chebyshev's Theorem states that for any dataset

• We have at least 75% of all the data within two deviations from the mean.

• We have at least 88.9% of all the data within three deviations from the mean.

• We have at least 93.8% of all the data within four deviations from the mean.

Or in general words "For any set of data (either population or sample) and for any constant k greater than 1, the proportion of the data that must lie within k standard deviations on either side of the mean is at least: $1-\frac{1}{k^2}$

Part a

For this case we can find the percentage required replaincg k =2 and we got:

$1-\frac{1}{k^2} =1- \frac{1}{2^2}= 1-0.25 = 0.75$

So we expected about 75% within two deviations from the mean

Part b

For this case we can find the percentage required replaincg k =2 and we got:

$1-\frac{1}{k^2} =1- \frac{1}{1.5^2}= 1-0.4444 = 0.556$

So we expected about 55.6% within 1.5 deviations from the mean

And the limits are:

$Lower = 3.41 -1.5*0.09 = 3.275$

$Upper = 3.41 +1.5*0.09 = 3.545$

Part c

We can calculate how many deviations we are within the mean with the limits with this formula:

$z =\frac{x-\mu}{\sigma}$

And using the lower limit we got:

$z = \frac{3.05-3.41}{0.09}=-4$

And with the upper limit we got:

$z = \frac{3.77-3.41}{0.09}=4$

So then the value of k =4 and the percentage is given by:

$1-\frac{1}{k^2} =1- \frac{1}{4^2}= 1-0.0625 = 0.9375$

3 0

3 years ago

How do you divide 3/8 by 6/7 in simplest form

Ipatiy [6.2K]

Answer:

7/16

Step-by-step explanation:

First, it is good to know that when you divide fractions, you are multiplying it's reciprocal.

6/7's reciprocal is 7/6, since the reciprocal is $\frac{1}{6/7}$

3/8 / 6/7 =

3/8 * 7/6

Now that we have that, we can just multiply the two:

3/8 * 7/6 =

21/48 =

7/16

Hope this helped :)

8 0

3 years ago

Math work pls help :)​

Math work pls help :)