5/-6+10i what is the answer to this

What is the perimeter of the figure on the coordinate plane below?

erik [133]

Answer:

26 units

Step-by-step explanation:

im not sure if this will help but i counted the spaces and added the halves

3 0

3 years ago

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IF U ANSWER THIS ILL GIVE BRAINLIST

mario62 [17]

Answer:

F

Step-by-step explanation:

That represents the equation correctly because 340 and filled (plus 5fl oz) and 650-8 ( subtract)

Hope this helps plz mark brainliest :D

8 0

3 years ago

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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

The ratio of boys to girls in the checkers club at one school is 2:3. the number of club members is more than 20 and fewer than

anastassius [24]

So our ratio is 2:3, therefore we are looking for a multiple of 2 and a multiple of 3 that are proportional to this ratio. We can start by multiplying the ratio by 2, which gets us 4:6, which, since 4+6 is ten, is not a correct answer. If you continue the pattern of multiplying the ration with different numbers, you would eventually discover that, if you multiply the ratio by 5, you get 10:15. Since 10+15 is equal to 25, we know that this is the correct answer

8 0

3 years ago

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Evaluate the expression when <br><br> a = -4 and c=5.<br><br> 6c-a

igomit [66]

=26
Explanation: 6 x 5 (c) = 30 30 - 4 (a) =26
Hope that helps :)

8 0

2 years ago

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