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densk [106]
3 years ago
9

A deep sea diving bell is being lowered at a constant rate. After 10 ​minutes, the bell is at a depth of 600 ft. After 40 minute

s the bell is at a depth of 1 comma 700 ft. What is the average rate of lowering per​ minute? Round to the nearest hundredth.
Mathematics
1 answer:
Degger [83]3 years ago
7 0
<h2>Average rate of lowering per​ minute = 36.67 ft/min</h2>

Step-by-step explanation:

After 10 ​minutes, the bell is at a depth of 600 ft

After 10 minutes

                   Depth of bell = 600 ft

After 40 minutes the bell is at a depth of 1,700 ft.

After 40 minutes

                   Depth of bell = 1700 ft

Time gap = 40 - 10 = 30 minutes

Depth fallen = 1700 - 600 = 1100 ft

We have

           Depth fallen = Average rate of lowering per​ minute x Time gap

           1100 = Average rate of lowering per​ minute x 30

           Average rate of lowering per​ minute = 36.67 ft/min

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The accompanying data contains the depth​ (in kilometers) and​ magnitude, measured using the Richter​ Scale, of all earthquakes
Sunny_sXe [5.5K]

Answer:

Depth:

μ =20.2025 km

M = 15.625 km

Range = 47.15 km

σ ≈ 15.92 km

Q₁ = 5.7375 km

Q₃ =  34.6675 km

Magnitude:

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

Step-by-step explanation:

The given data are;

Depth {}                                 Magnitude

0.76 {}                                    0.84

4.93 {}                                    0.47

8.16 {}                                     0.35

33.58 {}                                  1.32

21.2 {}                                     1.61

35.03 {}                                  4.57

10.05 {}                                   5.52

47.91 {}                                    1.99

For the Depth, we have;

The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km

The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;

0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;

(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km

The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km

The Standard deviation of, σ, is given as follows;

\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu  \right )^{2} }{N}}

Where;

x_i = The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 )

N = The total number of data point = 8

Substituting, (using Microsoft Excel) we get;

\sigma =\sqrt{\dfrac{\sum \left (x_i-20.2025  \right )^{2} }{8}} \approx 15.92 \ km

Q₁ = The first quartile = The (n + 1)/4th =  term arranged in increasing order

Q₁ = The (8 + 1)/4th term = The 2.25th term = 4.93 + (8.16 - 4.93)×0.25) = 5.7375 km

Q₃ = The first quartile = The 3×(n + 1)/4th =  term arranged in increasing order

Q₃ = The 3×(8 + 1)/4th term = The 6.75th term = 33.58 + 3×(35.03 - 33.58)×0.25) = 34.6675 km

For the magnitude, we have, using the same formulas and procedures as above;

μ = 2.08375

M = 1.465

Range, R = 5.17

σ = 1.801485 ≈ 1.8

Q₁ = 0.5625

Q₃ = 3.925

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Hard to select the correct answers without seeing the choices. Let's check a couple,


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