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3 years ago

Name an example of an acute angle, right angle, and straight angle according to their angle measures

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

8 0

Acute: anything less than 90 but greater than 0

so, m<CAD

right: angles at exactly 90 degrees, so m<AEC,

and obtuse angles are anything greater than 90 but less than 180 degrees,

so CDA (you can tell by the picture that it is bigger than 90- 90 degrees is exactly perpendicular)

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What is the solution for the determinant?

Liono4ka [1.6K]

<span>The value of the determinant of a 2x2 matrix is the product of the top-left and bottom-right terms minus the product of the top-right and bottom-left terms.

The value of the determinant of a 2x2 matrix is the product of the top-left and bottom-right terms minus the product of the top-right and bottom-left terms.

= [ (1)(-3)] - [ (7)(0) ]
= -3 - 0
= -3

Therefore, the determinant is -3.

Hope this helps!</span>

4 0

3 years ago

What is the answer to r - 4.5 < 11

Cloud [144]

R - 4.5 < 11
—
r would be 15.5
[ r = 15.5 ]

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2 years ago

Slove for y show all your work 12y + 4 = 8y-12

Vedmedyk [2.9K]

Answer:

y = -4

Step-by-step explanation:

12y + 4 = 8y-12

Subtract 8y from each side

12y - 8y +4 = -12

4y +4 = -12

Subtract 4 from each side

4y +4-4 = -12 -4

4y = -16

Divide by 4

4y/4 = -16/4

y = -4

8 0

3 years ago

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A forester studying the effects of fertilization on certain pine forests in the Southeast is interested in estimating the averag

ahrayia [7]

Answer:

$P(-2 \leq \bar X -\mu \leq 2)$

If we divide both sides by $\frac{\sigma}{\sqrt{n}}$ we got:

$P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})$

And we can use the normal distribution table or excel to find the probabilites and we got:

$P(-1.5 \leq Z \leq 1.5)= P(Z$ Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the area of a population, and for this case we know the distribution for X is given by:

$X \sim N(M,4)$

Where $\mu=M$ and $\sigma=4$

We select a a sample of n =4 and since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want to find this probability:

$P(-2 \leq \bar X -\mu \leq 2)$

If we divide both sides by $\frac{\sigma}{\sqrt{n}}$ we got:

$P(\frac{-2}{\frac{4}{\sqrt{9}}}\leq Z \leq \frac{2}{\frac{4}{\sqrt{9}}})$

And we can use the normal distribution table or excel to find the probabilites and we got:

$P(-1.5 \leq Z \leq 1.5)= P(Z$

8 0

3 years ago

(8Q) Tell whether the function exhibits damped oscillation. If it does, identify the damping factor and tell whether the damping

dezoksy [38]

Answer:

Option c.

No damping

Step-by-step explanation:

We can easily solve this question by using a graphing calculator or any plotting tool.

The function is

f(x) = (√11)*cos(3.7x)

Which can be seen in the picture below

We can notice that f(x) is a cosine with maximum amplitude of (√11). Neither this factor nor the multiplication of x by 3.7 serve as a damping factor since they are constants.

f(x) does not present any dampening

5 0

3 years ago

Read 2 more answers