Answer:
B
Explanation:
Adding a dopant is correct on edge.
Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).
Answer:
Explanation:
Theorem of Binomial Distribution will apply here.
n = 29 , p = .67 , q = 0.33
mean = np = 29 x .67 = 19.43
Standard Deviation = √npq
= √29 x .67 x .33
= √6.4
= 2.53
=
Answer:

Explanation:
The change in electrical potential energy of a charged particle moving through a potential difference is given by

where
q is the magnitude of the charge of the particle
is the potential difference
In this problem:
- the charge of the particle is 3.00 elementary charges, so

- the potential difference is

So, the change in electrical potential energy is
