<span>Position at t=0.35s is 0.2 m
Velocity at t = 0.35s is -0.2 m/s
Since this is college level mathematics, the use of the word "acceleration" should indicate to you that you've been given the 2nd derivative of a function specifying the location of point a. And since you've been asked for the velocity, you know that you want the 1st derivative of the function. And since you've also been asked for the position, you also want the function itself. So let's calculate the desired anti-derivatives.
f''(t) = -1.08 sin(kt) - 1.44 cos(kt)
The integral of f''(t) with respect to t is:
f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k + C
In order to find out what C is, we know that at time t=0, v = 0.36 m/s. So let's plug in the values and see what C is:
f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k + C
0.36 = (1.08 cos(3*0) - 1.44 sin(3*0))/3 + C
0.36 = (1.08 cos(0) - 1.44 sin(0))/3 + C
0.36 = (1.08*1 - 1.44*0)/3 + C
0.36 = 0.36 + C
0 = C
So the first derivative will be f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k
Now to get the actual function by integrating again. Giving:
f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 + C
And let's determine what C is:
f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2 + C
0.16 = (1.08 sin(3*0) + 1.44 cos(3*0))/3^2 + C
0.16 = (1.08 sin(0) + 1.44 cos(0))/9 + C
0.16 = (1.08*0 + 1.44*1)/9 + C
0.16 = 1.44/9 + C
0.16 = 0.16 + C
0 = C
So C = 0 and the position function is: f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2
So now, we can use out position and velocity functions to get the desired answer:
Position:
f(t) = (1.08 sin(kt) + 1.44 cos(kt))/k^2
f(t) = (1.08 sin(3*0.35) + 1.44 cos(3*0.35))/3^2
f(t) = (1.08 sin(1.05) + 1.44 cos(1.05))/9
f(t) = (1.08*0.867423226 + 1.44*0.497571048)/9
f(t) = (0.936817084 + 0.716502309)/9
f(t) = 1.653319393/9
f(t) = 0.183702155
So the position of point a at t=0.35s is 0.2 m
Now for the velocity:
f'(t) = (1.08 cos(kt) - 1.44 sin(kt))/k
f'(t) = (1.08 cos(3*0.35) - 1.44 sin(3*0.35))/3
f'(t) = (1.08 cos(1.05) - 1.44 sin(1.05))/3
f'(t) = (1.08*0.497571048 - 1.44*0.867423226)/3
f'(t) = (0.537376732 - 1.249089445)/3
f'(t) = -0.711712713/3
f'(t) = -0.237237571
So the velocity at t = 0.35s is -0.2 m/s</span>
Answer:
The sample consisting of 64 data values would give a greater precision.
Step-by-step explanation:
The width of a (1 - <em>α</em>)% confidence interval for population mean μ is:
So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (<em>n</em>).
That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.
Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.
The two sample sizes are:
<em>n</em>₁ = 25
<em>n</em>₂ = 64
The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.
Width for n = 25:
![\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{25}}=\frac{1}{5}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]](https://tex.z-dn.net/?f=%5Ctext%7BWidth%7D%3D2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7B25%7D%7D%3D%5Cfrac%7B1%7D%7B5%7D%5Ccdot%20%5B2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Csigma%5D)
Width for n = 64:
![\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\cdot [2\cdot z_{\alpha/2}\cdot \sigma]](https://tex.z-dn.net/?f=%5Ctext%7BWidth%7D%3D2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7B64%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5Ccdot%20%5B2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Csigma%5D)
Thus, the sample consisting of 64 data values would give a greater precision
Start with the solution
x=-25
use the reverse of what they asked
they wanted division and addition
we use multiplication and subtract
ok
x=-25
we can use subtraction and multiplication in any order
minus 10 from both sides
x-10=-35
multiply both sides by 4
4(x-10)=-140
4x-40=-140 is a possible equation
<h3>
Answer:</h3>
- using y = x, the error is about 0.1812
- using y = (x -π/4 +1)/√2, the error is about 0.02620
<h3>
Step-by-step explanation:</h3>
The actual value of sin(π/3) is (√3)/2 ≈ 0.86602540.
If the sine function is approximated by y=x (no error at x = 0), then the error at x=π/3 is ...
... x -sin(x) @ x=π/3
... π/3 -(√3)/2 ≈ 0.18117215 ≈ 0.1812
You know right away this is a bad approximation, because the approximate value is π/3 ≈ 1.04719755, a value greater than 1. The range of the sine function is [-1, 1] so there will be no values greater than 1.
___
If the sine function is approximated by y=(x+1-π/4)/√2 (no error at x=π/4), then the error at x=π/3 is ...
... (x+1-π/4)/√2 -sin(x) @ x=π/3
... (π/12 +1)/√2 -(√3)/2 ≈ 0.026201500 ≈ 0.02620
Answer:
He can make 6 groups
Step-by-step explanation:
To solve this problem first you need to find the GCF or greatest common factor. Now you can put 5 chocolate chip cookies in each group, 3 peanut butter, and 4 sugar cookies. These are all relatively prime so that would be your final answer.
