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3 years ago

john has 8 pencils. tim has 2 less pencils than john. which expression shows how many pencils tim has?

Mathematics

2 answers:

Misha Larkins [42]3 years ago

7 0

Answer:

Step-by-step explanation:

cuz i said

enot [183]3 years ago

4 0

Answer:

Step-by-step explanation:

8-2 would be the amount of pencils tim has

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Lisa has 2 red dresses, 3 green dresses and x blue dresses hanging in her closet. If lisa chooses one dress at random, the proba

Vladimir79 [104]

Answer:

the dess was actually blue and black

Step-by-step explanation:

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3 years ago

A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 5.2 on a placement test. What is

Otrada [13]

Answer:

29.5+/-1.11

= ( 28.39, 30.61)

Therefore, the 90% confidence interval (a,b) =( 28.39, 30.61)

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 29.5

Standard deviation r = 5.2

Number of samples n = 59

Confidence interval = 90%

z-value (at 90% confidence) = 1.645

Substituting the values we have;

29.5+/-1.645(5.2/√59)

29.5+/-1.645(0.676982337100)

29.5+/-1.113635944529

29.5+/-1.11

= ( 28.39, 30.61)

Therefore, the 90% confidence interval (a,b) =( 28.39, 30.61)

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2 years ago

There are 3 sides in a right triangle. The two legs and what? The answer is one word and begins with "h".

V125BC [204]

Hypotenuse, you can find it by using the <span>Pythagorean theorem</span>

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3 years ago

What is the value of 8+23×5?

Leto [7]

Answer: 123

Step-by-step explanation:

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3 years ago

Read 2 more answers

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