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Soloha48 [4]
3 years ago
5

To evaluate the integrity of underground water lines, the department of public works randomly selects 20 sites in the city to un

earth and observe the water lines. At 5 of the sites, the water lines needed repair. The department of public works concludes that about one-fourth of underground water lines throughout the city need repair.
3. To award prizes at a hockey game, four tickets with individual seat
numbers printed on them are picked from a barrel. Since Elvis's
section was not selected for any of the four prizes, he assumes that
they forgot to include the entire section in the drawing.
Mathematics
2 answers:
ollegr [7]3 years ago
6 0
4. Normal body temperature is 98.6 degrees Fahrenheit, although it varies slightly for some people. Someone ?
Zanzabum3 years ago
6 0
98.6, I think, tell me how you do!
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7-14m=2m=5 please help​
Firdavs [7]

Answer:

M=1/6

Step-by-step explanation:

6 0
3 years ago
L'manbergian come to this and say hi
Vikki [24]

Answer:

Hey

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
3<br> Given the function f (x) = 3.84 – 5x2 + 2x – 3, evaluate for f(
Artemon [7]

Step-by-step explanation:

f(x) = 3x⁴ - 5x² + 2x - 3

When x = -1,

f(-1) = 3(-1)⁴ - 5(-1)² + 2(-1) - 3 = 3 - 5 - 2 - 3 = -7.

3 0
3 years ago
Una caja de 25 newtons se suspende mediante un cable con diámetro de 2cm¿cuál es el esfuerzo aplicado al cable?
Naddik [55]

El cable experimenta un esfuerzo axial de 79577.472 pascales por el peso de la caja.

<h3>¿Cómo calcular el esfuerzo aplicado sobre el cable?</h3>

La caja tiene masa y está sometida a un campo gravitacional, por tanto, tiene un peso (W), en newtons. Por el principio de acción y reacción (tercera ley de Newton), encontramos que el cable es tensionado debido a ese peso y su área transversal experimenta un esfuerzo axial (σ), en pascales.

Asumiendo una distribución uniforme de la fuerza sobre toda la superficie transversal de la cuerda, tenemos que el esfuerzo axial se calcula mediante la siguiente expresión:

σ = W / (π · D² / 4)

Donde:

  • W - Peso de la caja, en newtons.
  • D - Diámetro del área transversal de la caja, en metros.

Si sabemos que W = 25 N y D = 0.02 m, entonces el esfuerzo axial aplicado a la cuerda es:

σ = 25 N / [π · (0.02 m)² / 4]

σ ≈ 79577.472 Pa

<h3>Observación</h3>

La falta de problemas verificados en español sobre esfuerzos axiales obliga a buscar uno equivalente en inglés.

Para aprender más sobre esfuerzos axiales: brainly.com/question/13683145

#SPJ1

5 0
1 year ago
7 divided 2/3 , give answer in its simplest form
cluponka [151]

Answer:

21/2 or 10 1/2

8 0
3 years ago
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