Find the product. -2x(x2

Which reason justifies the following statement?

amid [387]

Answer:

(A) Complements of the same angle are congruent.

Step-by-step explanation:

Since, ∠CEF is complementary to ∠DCF, therefore

∠CEF+∠DCF=90° (1)

Also, ∠CEG=∠CEF+∠FEG

⇒90°=∠CEF+∠FEG (2)

Therefore, from(1) and (2), we get

∠CEF+∠DCF=90°

∠CEF+∠FEG=90°

Subtracting both the equations, we get

∠DCF-∠FEG=0

∠DCF=∠FEG.

Hence, ∠DCF≅∠FEG by the rule: Complements of the same angle are congruent.

3 0

3 years ago

Read 2 more answers

What's measure E & D?

Kipish [7]

You'll have to show work guaranteed.......so i'll tell you that you need to inverse

3 0

3 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

"Find the value of the derivative (if it exists) at each indicated extremum. (If an answer does not exist, enter DNE.) f (x) = c

Oliga [24]

$\bf f(x)=cos\left( \frac{\pi x}{2} \right)\implies \cfrac{df}{dx}=\stackrel{chain~rule}{-sin\left( \frac{\pi x}{2} \right)\cdot \frac{\pi }{2}}$

now, what is the value of the derivative at any extrema on the original function? well, by definition, if the original function has an extrema anywhere, the derivative will be the derivative of a horizontal tangent line to it, and by definition it'll be 0.

8 0

2 years ago

Find the distance between M(-2,3) and N (8,2)

In-s [12.5K]

1 and 6................. hope I helped

4 0

3 years ago

Find the product. -2x(x2 - 3)