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posledela
4 years ago
11

Use the temperatures box-and-whisker plot to solve. What are the quartiles of the data? A. lower quartile (Q1) = 14 median (Q2)

= 22 upper quartile (Q3) = 27 B. lower quartile (Q1) = 12 median (Q2) = 22 upper quartile (Q3) = 30 C. lower quartile (Q1) = 14 median (Q2) = 20 upper quartile (Q3) = 27 D. lower quartile (Q1) = 14 median (Q2) = 22 upper quartile (Q3) = 26
Physics
2 answers:
blondinia [14]4 years ago
6 0
Use the temperatures box-and-whisker plot to solve. What are the quartiles of the data? D. Is the correct answer, Hope I helped, and good luck!
Semenov [28]4 years ago
4 0
 D. lower quartile (Q1) = 14 median (Q2) = 22 upper quartile (Q3) = 26
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Distance Between Two Cars A car leaves an intersection traveling west. Its position 4 sec later is 19 ft from the intersection.
faltersainse [42]

Answer:

The answer to the question is;

The rate at which the distance between the two cars is changing is equal to 14.4 ft/sec.

Explanation:

We note that the distance  traveled by each car after 4 seconds is

Car A = 19 ft in the west direction.

Car B = 26 ft in the north direction

The distance between the two cars is given by the length of the hypotenuse side of a right angled triangle with the north being the y coordinate and the  west being the x coordinate.

Therefore, let the distance between the two cars be s

we have

s² = x² + y²

= (19 ft)² + (26 ft)² = 1037 ft²

s = \sqrt{1037 ft^2} = 32.202 ft.

The rate of change of the distance from their location 4 seconds after they commenced their journeys is given by;

Since s² = x² + y² we have

\frac{ds^{2} }{dt} = \frac{dx^{2} }{dt}  + \frac{dy^{2} }{dt}

→ 2s\frac{ds }{dt} = 2x\frac{dx}{dt}  + 2y\frac{dy }{dt} which gives

s\frac{ds }{dt} = x\frac{dx}{dt}  + y\frac{dy }{dt}

We note that the speeds of the cars were given as

Car B moving north = 12 ft/sec, which is the y direction and

Car A moving west = 8 ft/sec which is the x direction.

Therefore

\frac{dy }{dt} =  12 ft/sec and

\frac{dx}{dt} = 8 ft/sec

s\frac{ds }{dt} = x\frac{dx}{dt}  + y\frac{dy }{dt} becomes

32.202 ft.\frac{ds }{dt} = 19 ft \times 8 \frac{ft}{sec}  + 26ft\times 12\frac{ft}{sec}  = 464 ft²/sec

\frac{ds }{dt} = \frac{464\frac{ft^{2} }{sec} }{32.202 ft.} = 14.409 ft/sec ≈ 14.4 ft/sec to one place of decimal.

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3 years ago
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Explanation:

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where h is height,p is density, g is 10m/s²

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P=6.8x10³ Hgmm

3 0
3 years ago
Calculate the density of each ball. Use the formula
Paraphin [41]

Answer:

⇒ 0.07 g/cm3  

⇒ 1.15 g/cm3

Explanation:

3 0
3 years ago
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What is the elastic potential energy of a spring that is compressed a distance of 0.35 m and has a spring constant of 71.8 N/m?
Vitek1552 [10]

Answer:

P.E = 4.398 Joules.

Explanation:

Given the following data;

Spring constant, k = 71.8N/m

Displacement, x = 0.35m

To find the elastic potential energy;

The elastic potential energy of an object is given by the formula;

P.E = \frac {1}{2}kx^{2}

Substituting into the equation, we have;

P.E = \frac {1}{2}*71.8 *(0.35)^{2}

P.E = 35.9 * 0.1225

Elastic potential energy = 4.398 Joules.

<em>Therefore, the elastic potential energy of the spring is 4.398 Joules. </em>

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Vilka [71]
What do u mean by that

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