F = k / (r^2)
dF/dt = -2k / (r^3) dr/dt
When r = 10^4 km and dr/dt = 0.2 km/sec
dF/dt = -2(10)^3 (0.2) / ((10^4)^3) = -400 / 10^7 = -0.00004 N/s
The gravitational force is changing at the rate of -0.00004 N/s
Answer:
C. ± 2.326 years.
Step-by-step explanation:
We have the standard deviation for the sample. So we use the t-distribution to solve this question.
We have that to find our 
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
Now, we have to find z in the Ztable as such z has a pvalue of 
.
So it is z with a pvalue of 
, so ![z = 2.326/tex]Now, find the width of the interval[tex]W = z*\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=z%20%3D%202.326%2Ftex%5D%3C%2Fp%3E%3Cp%3E%3Cstrong%3ENow%3C%2Fstrong%3E%2C%20find%20the%20width%20of%20the%20interval%3C%2Fp%3E%3Cp%3E%5Btex%5DW%20%3D%20z%2A%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In which 
is the standard deviation of the population and n is the size of the sample.
In this question:
So
The correct answer is:
C. ± 2.326 years.
Answer:
3 1/2
Step-by-step explanation:
So because the fraction in attatched to the whole numbers have the same denominator you can literally just subtract the whole numbers and the fractions. (You do them seperatly) (5-2)=3 (3/4-1/4) = 2/4 or 1/2
Answer:
First and second
Step-by-step explanation:
Answer:
yes
Step-by-step explanation:
A relation is a function in which for each input there is only one output.
In a relation, y is a function of x.
Now we look at the mapping
For input 6, -1 is the output
for input -1, 2 is the output
for input 4, 3 is the output
for input 0, 3 is the output
For each input , there is an output. Input is not repeating. Input is occurring only once.
So this relation is a function.
