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Reil [10]
3 years ago
9

GIVING BRAINLIEST how many O's are in one thousand

Mathematics
2 answers:
Jlenok [28]3 years ago
8 0

Answer:

3

Step-by-step explanation:

Airida [17]3 years ago
4 0
2 because One thOusand
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A company installs 5,000 light bulbs. the lifetimes of the lightbulbs are approximately normally distributed with a mean of 500
Sergio039 [100]
First, determine the z-score of 675. 
                               z = (675 - 500) / 100 = 1.75
The z-score of 500 is,
                               z = 0. 
Subtracting the z-scores will give us 1.75. This is equal to 0.9599. 
                        = 0.9599 - 0.5 = 0.4599
Multiplying this to the given number of light bulbs,
                             n = 0.4599 x 5000 = 2299.5
Therefore, there is approximately 2300 light bulbs expected to last between 500 to 675 hours.
5 0
3 years ago
The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f
WITCHER [35]

Answer:

Part 1) A=36(\pi-2)\ cm^2

Part 2) P=6(\pi+2\sqrt{2})\ cm

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

so

A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)

we have that the base and the height of triangle is equal to the radius of the circle

r=12\ cm\\b=12\ cm\\h=12\ cm

substitute

A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2

simplify

Factor 36

A=36(\pi-2)\ cm^2

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r

substitute the given values

C=\frac{1}{2}\pi (12)\\C=6\pi\ cm

Applying the Pythagorean Theorem

The hypotenuse of right triangle is equal to

AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm

simplify

AC=12\sqrt{2}\ cm

Find the perimeter

P=(6\pi+12\sqrt{2})\ cm

simplify

Factor 6

P=6(\pi+2\sqrt{2})\ cm

4 0
3 years ago
<img src="https://tex.z-dn.net/?f=5%20%2B%204%20-%20%28%205%20-%20%286%20-%205%29%29" id="TexFormula1" title="5 + 4 - ( 5 - (6 -
Ket [755]
(6-5)=1 (5-1)=4 5+4=9 9-4=5
8 0
3 years ago
Read 2 more answers
Help me plzz.....................
Ierofanga [76]
This is Transitive Property. 

Blame me if you get it wrong :\
6 0
3 years ago
What is the resulting equation when the expression for y in the second equation is substituted into the first equation-3x+2y=-3
Alex

Answer:

x=-5/11, y=-9/11. (-5/11, -9/11).

Step-by-step explanation:

3x+2y=-3

y=4x+1

---------------

3x+2(4x+1)=-3

3x+8x+2=-3

11x=-3-2

11x=-5

x=-5/11

y=4(-5/11)+1

y=-20/11+11/11

y=-9/11

5 0
3 years ago
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