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GalinKa [24]
2 years ago
5

An Internet service provider sampled 540 customers, and finds that 75 of them experienced an interruption in high-speed service

during the previous month.a) Obtain a point estimate, to at least three decimal places, for the population proportion of all customers who experienced an interruption in high-speed service during the previous month.b) Assuming the sample size is less than 5% of the population size, why can the sampling distribution of p-hat, the sample proportion of customers who experienced an interruption in high-speed service during the previous month, be approximately normal?c) Construct a 90% confidence interval for the proportion of all customers who experienced an interruption in high-speed service during the previous month.d) You wish to conduct your own study to determine the proportion of all customers who experienced an interruption in high-speed service during the previous month. What sample size would be needed for the estimate to be within 2 percentage points with 98% confidence if you use the point estimate obtained in part (a)?e) You wish to conduct your own study to determine the proportion of all customers who experienced an interruption in high-speed service during the previous month. What sample size would be needed for the estimate to be within 2 percentage points with 98% confidence if you don
Mathematics
1 answer:
Nastasia [14]2 years ago
7 0

Answer:

ill answer in a minute

Step-by-step explanation:

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d=54-29

d=25

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Please solve this for me
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Brainilest!

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V = (15 x 28) / 3

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The equation t^3=a^2 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
ankoles [38]

Answer:

     2√2

Step-by-step explanation:

We can find the relationship of interest by solving the given equation for A, the mean distance.

<h3>Solve for A</h3>

  T^3=A^2\\\\A=\sqrt{T^3}=T\sqrt{T}\qquad\text{take the square root}

<h3>Substitute values</h3>

The mean distance of planet X is found in terms of its period to be ...

  D_x=T_x\sqrt{T_x}

The mean distance of planet Y can be found using the given relation ...

  T_y=2T_x\\\\D_y=T_y\sqrt{T_y}=2T_x\sqrt{2T_x}=(2\sqrt{2})T_x\sqrt{T_x}\\\\D_y=2\sqrt{2}\cdot D_x

The mean distance of planet Y is increased from that of planet X by the factor ...

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8 0
2 years ago
Can I get help with this please
drek231 [11]

Answer:

3)  1 5/6 mi

4)  a. 4 cm, 6 ft

    b. 6.4 cm, 9.6 ft

    c. same as part a

Step-by-step explanation:

3) Each of the given distances appears twice in the sum of side measures that is the perimeter. Hence by walking the perimeter twice, Kyle walks each of the given distances 4 times. His total walk is ...

  4×1/3 + 4×1/8 = 4/3 + 4/8

  = 1 1/3 + 1/2 = 1 2/6 + 3/6

  = 1 5/6 . . . . . miles

___

4) Since the figure is rectilinear (all angles are right angles, and all sides are straight lines), the sum of partial dimensions in one direction is equal to the whole dimension in that direction.

a. 8 cm = 4 cm + x

  8 cm - 4 cm = x = 4 cm

The distance in the room is ...

  (4 cm)×(1.5 ft/cm) = 6 ft

b. 10.3 cm = 3.9 cm + y

  10.3 cm - 3.9 cm = y = 6.4 cm

The distance in the room is ...

  (6.4 cm)×(1.5 ft/cm) = 9.6 ft

c. The answer to part b was obtained in the same way as the answer to part a.  The unknown dimension is the difference of given dimensions. The actual length in the room is the model length multiplied by the inverse of the scale factor.

3 0
3 years ago
Please help!
pishuonlain [190]

Answer:

0.03125 inches 1 hour is the anwser

this also involves the conversion factor

hope this helps!!

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