Ask question

Ask question

All categories

2 years ago

14

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar. In a survey conduc

ted in an office, 60% of the sample respondents agreed that employees should have flexible working hours. If the margin of error is ±2%, the expected population proportion that wants flexible working hours is between __% and __%

1 answer:

jarptica [38.1K]2 years ago

3 0

Answer: The expected population that wants flexible working hours in ranging from 58% to 62 %.

Explanation:

Percentage of sample respondents agreed for flexible working hours = 60%

The margin of the error = ± 2%

The expected population that wants flexible working hours in ranging from:

$\text{percentage of agreed}-2\%$ lower range

$\text{percentage of agreed}+\2%$ upper range

$60\%-2\% to 60\%+2\%$ that is 58% to 62 %.

The expected population that wants flexible working hours in ranging from 58% to 62 %.

You might be interested in

A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from

Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat ( $U_{g}$ ), in joules, is equal to the sum of the final translational kinetic energy ( $K$ ), in joules, and work losses due to air resistance ( $W_{l}$ ), in joules:

$U_{g} = K +W_{l}$ (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

$m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l}$ (2)

Where:

$m$ - Mass of the cat, in kilograms.

$g$ - Gravitational acceleration, in meters per square second.

$h$ - Initial height of the cat, in meters.

$v$ - Final speed of the cat, in meters per second.

If we know that $m = 2.70\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h = 5.20\,m$ and $W_{l} = 120\,J$ , then the final speed of the cat is:

$v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }$

$v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }$

$v \approx 7.586\,\frac{m}{s}$

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0

2 years ago

Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68

den301095 [7]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, $y_0=1500\ m$

Initial horizontal velocity is, $u_x=688\ m/s$

Initial vertical velocity is, $u_y=0(\textrm{Horizontal velocity only initially)}$

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is $S=y-y_0=0-1500=-1500\ m$

Acceleration in the vertical direction is due to gravity, $g=-9.8\ m/s^2$

Therefore, the displacement of the bomb is given as:

$S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s$

So, the bomb will remain in air for 17.5 s before hitting the ground.

6 0

2 years ago

Write a hypothesis about how the force applied to a cart affects the acceleration of the cart. Use the "if . . . then . . .becau

Lesechka [4]

"If one increases the force on an object, its acceleration increases too because the push it feels is greater"

We have the 2nd law of Newton that relates the 3 concepts; F=m*a. We have that if the mass of an object increases (put weight in luggage), the accelearation decreases; in fact it is inversely proportional to the mass. Hence if the mass is doubled, acceleration is halved. Accelerations is proportional to force; if one doubles the force, the acceleration doubles too.

7 0

2 years ago

Read 2 more answers

We know that the earth's axis is tilted 23 ½ degrees. On or about June 21 or 22 each year, the summer solstice occurs for those

Dima020 [189]

<span>D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.</span>

8 0

2 years ago

Read 2 more answers

A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900

Whitepunk [10]

It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.

5 0

1 year ago

Read 2 more answers