Answer:
Standard deviation = 3
Explanation:
Given


Required
Determine the standard deviation
First, we need to determine the variance;

This gives:



Know that:

Where SD represents standard deviation
This gives

Take square root


Answer:
<em><u>172,000 second </u></em>
<em><u>I'M</u></em><em><u> </u></em><em><u>NOT</u></em><em><u> </u></em><em><u>SURE</u></em><em><u> </u></em><em><u>THAT</u></em><em><u> </u></em><em><u>THIS</u></em><em><u> </u></em><em><u>IS</u></em><em><u> </u></em><em><u>RIGHT</u></em><em><u> </u></em><em><u>OR</u></em><em><u> </u></em><em><u>WRONG</u></em><em><u> </u></em><em><u> </u></em><em><u>IF</u></em><em><u> </u></em><em><u>IT'S</u></em><em><u> </u></em><em><u>WRONG</u></em><em><u> </u></em><em><u>THEN</u></em><em><u> </u></em><em><u>SORRY</u></em><em><u> </u></em>
Only within the same technology. / / /
If both of the bulbs you're comparing are incandescent, or both fluorescent, or both CFL, or both LED, then the one that uses more power is brighter. But a CFL with the same brightness as an incandescent bulb uses less power, and an LED bulb with the same brightness as both of those uses less power than either of them.
Answer:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.