Answer:
we reject H₀
Step-by-step explanation: Se annex
The test is one tail-test (greater than)
Using α = 0,05 (critical value ) from z- table we get
z(c) = 1,64
And Test hypothesis is:
H₀ Null hypothesis μ = μ₀
Hₐ Alternate hypothesis μ > μ₀
Which we need to compare with z(s) = 2,19 (from problem statement)
The annex shows z(c), z(s), rejection and acceptance regions, and as we can see z(s) > z(c) and it is in the rejection region
So base on our drawing we will reject H₀
Step-by-step explanation:
x² + (y − 1)² = 9
This is a circle with center (0, 1) and radius 3. We can parameterize it using sine and cosine.
Use the starting point to determine which should be sine and which should be cosine.
Use the direction to determine the signs.
Use the number of revolutions and the interval to determine coefficient of t.
(A) Once around clockwise, starting at (3, 1). 0 ≤ t ≤ 2π.
The particle starts at (3, 1), which is 0 radians on a unit circle. It makes 1 revolution (2π radians). Therefore:
x = 3 cos t
y = 1 − 3 sin t
(B) Two times around counterclockwise, starting at (3, 1). 0 ≤ t ≤ 4π.
The particle starts at (3, 1), which is 0 radians on a unit circle. It makes 2 revolutions (4π radians). Therefore:
x = 3 cos t
y = 1 + 3 sin t
(C) Halfway around counterclockwise, starting at (0, 4). 0 ≤ t ≤ π.
The particle starts at (0, 4), which is π/2 radians on a unit circle. It makes 1/2 revolution (π radians). Therefore:
x = -3 sin t
y = 1 + 3 cos t
Hello,
A=πR²=3.141592...*15²=706,8583470...( ft²)
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