Answer:
0.049168726 light-years
Step-by-step explanation:
The apparent brightness of a star is
where
<em>L = luminosity of the star (related to the Sun)
</em>
<em>d = distance in ly (light-years)
</em>
The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.
Hence the apparent brightness of Alpha Centauri A is
According to the inverse square law for light intensity
where
light intensity at distance 
light intensity at distance 
Let be the distance we would have to place the 50-watt bulb, then replacing in the formula
Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.
Answer:
Test statistic = - 0.851063
- 2.520463
Step-by-step explanation:
H0 : μ ≥ 15
H1 : μ < 15
Sample mean, xbar = 14.5
Sample standard deviation, s = 4.7
Sample size = 64
Teat statistic :
(xbar - μ) ÷ (s/√(n))
(14.5 - 15) ÷ (4.7/√(64))
= - 0.851063
The critical value at α = 0.05
Using the T - distribution :
Degree of freedom, df = 64 - 1 = 63
Tcritical(0.05, 63) = 1.6694
Test statistic - critical value
-0.851063 - 1.6694
= - 2.520463
Answer:
Below.
Step-by-step explanation:
1. a^2 - b^2 = (a - b)(a + b)
a^2 and b^2 are 2 perfect squares and a^2 - b^2 are their difference.
2. The order of multiplication does not matter because in algebra multiplication is commutative ( meaning any order gives the same result).
