The car costs $7500
7500-3000=4500 to be paid in monthly payments
4500/36=125
Ryan paid 125 per month
The option D is the correct option.
According to the statement
we have to define the sample variance in the words.
So,
The sample variance is the sum of the squared deviations from the mean divided by the number of measurements minus one.
and in the case of the population variance it becomes
The population variance is the average of the squared distances of the measurements on all units in the population from the mean.
This is the method by which we define the sample variance and population variance.
So, The option D is the correct option.
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Answer:
a. v(t)= -6.78
+ 16.33 b. 16.33 m/s
Step-by-step explanation:
The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=
=
. We now multiply both sides of the equation by the integrating factor.
μv' + μkv = μg ⇒ v' + kv = g ⇒ [v]' = g. Integrating, we have
∫ [v]' = ∫g
v = 
+ c
v(t)= + c
.
From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have
9.55 = 9.8 × 15/9 + c
= 16.33 + c
c = 9.55 -16.33 = -6.78.
So, v(t)= 16.33 - 6.78. m/s = - 6.78 + 16.33 m/s
b. Velocity of object at time t = 0.5
At t = 0.5, v = - 6.78
+ 16.33 m/s = 16.328 m/s ≅ 16.33 m/s
Rearranging y = 8x -4y = 12 in the y = mx + b form gives y = 2x - 3
y = 2x - 3
y = mx - 6 (already in y = mx + b form)
--------------
Parallel lines have the same slope.
m = 2
Note that the two lines have the different y-intercepts and are not the same line.
We know that
At sea level, the height is 0 and
the pressure is 98 kilopascals
At 1000 ft, the height is 1000 and
the pressure <span>decreases about 11.41%
</span>i(100%-11.41%)/100----> (0.8859)
therefore
1) at 1000 ft--------> the pressure is 98*(0.8859)----------> 86.82 kilopascals
2) at 2000 ft--------> the pressure is 86.82*(0.8859)-------> 76.91 kilopascals
3) at 3000 ft--------> the pressure is 76.91*(0.8859)-------> 68.14 kilopascals
4) at 4000 ft---------> the pressure is 68.14*(0.8859)-------> 60.36 kilopascals
the answer is
<span>The pressure at an altitude of 4000 m is about </span>60.36 kilopascals
