Find the surface area Of the figure. 9 cm 1 cm Insert Answer 5 cm 5 cm 1 cm

<img src="https://tex.z-dn.net/?f=4%5Csqrt%7B3%7D%2B%5Csqrt%7B3%7D" id="TexFormula1" title="4\sqrt{3}+\sqrt{3}" alt="4\sqrt{3}+\

Firlakuza [10]

Answer:

it's 5\sqrt{3}[/tex]

Step-by-step explanation:

hope it helps you

5 0

2 years ago

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Line I is parallel to line M. If the measure of 6 is 75 degrees, what is the measure of 3?

Doss [256]

In the given diagram, the measure of ∠3 will be 105°.

In the given diagram, ∠3 and ∠6 are consecutive interior angles.

<h3>How to form supplementary angles by transversal?</h3>

If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary.

That is,

∠3 + ∠6 = 180°

From the given information,

∠6 = 75°

Then,

∠3 + 75° = 180°

∠3 = 180° - 75°

∠3 = 105°

Hence, the measure of ∠3 will be 105°.

Learn more about the measures of angles here: brainly.com/question/2883630

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6 0

1 year ago

an outlier may be defined as a data point that is more than 1.5 times the interquartile range below the lower quartile or is mor

Ira Lisetskai [31]

Supposing a normal distribution, we find that:

The diameter of the smallest tree that is an outlier is of 16.36 inches.

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We suppose that tree diameters are normally distributed with mean 8.8 inches and standard deviation 2.8 inches.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

Mean of 8.8 inches, thus $\mu = 8.8$ .
Standard deviation of 2.8 inches, thus $\sigma = 2.8$ .

The interquartile range(IQR) is the difference between the 75th and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = -0.675(2.8)$

$X = 6.91$

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

$Z = \frac{X - \mu}{\sigma}$

$0.675 = \frac{X - 8.8}{2.8}$

$X - 8.8 = 0.675(2.8)$

$X = 10.69$

The IQR is:

$IQR = 10.69 - 6.91 = 3.78$

What is the diameter, in inches, of the smallest tree that is an outlier?

The diameter is 1.5IQR above the 75th percentile, thus:

$10.69 + 1.5(3.78) = 16.36$

The diameter of the smallest tree that is an outlier is of 16.36 inches.

A similar problem is given at brainly.com/question/15683591

3 0

2 years ago

Emma bought three cans of tennis balls for $4 each. Sales tax for all three cans was $0.95. Wh

LiRa [457]

Answer:

4(3)+0.95=X

Step-by-step explanation:

6 0

2 years ago

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Given that f(x) = 19x2 + 152, solve the equation f(x) = 0

telo118 [61]

Option A

The solution is:

$x = \pm 2i \sqrt{2}$

Solution:

$f(x) = 19x^2+152$

We have to solve the equation f(x) = 0

Let f(x) = 0

$0=19x^2+152$

Solve the above equation

$19x^2 + 152 = 0$

$\mathrm{Subtract\:}152\mathrm{\:from\:both\:sides}\\\\19x^2+152-152=0-152\\\\Simplify\ the\ above\ equation\\\\19x^2 = -152\\\\\mathrm{Divide\:both\:sides\:by\:}19\\\\\frac{19x^2}{19} = \frac{-152}{19}\\\\x^2 = -8$

Take square root on both sides

$x = \pm \sqrt{-8}\\\\x = \pm \sqrt{-1}\sqrt{8}\\\\\mathrm{Apply\:imaginary\:number\:rule}:\quad \sqrt{-1}=i\\\\x = \pm i\sqrt{8}\\\\x = \pm i \sqrt{2 \times 2 \times 2}\\\\x = \pm 2i\sqrt{2}$

Thus the solution is found

5 0

2 years ago

Find the surface area Of the figure. 9 cm 1 cm Insert Answer 5 cm 5 cm 1 cm ​

Find the surface area Of the figure. 9 cm 1 cm Insert Answer 5 cm 5 cm 1 cm