Answer: 3 cofficients are there ( 5, 2, and 6).
Answer:
This means that after going around the unit circle once (2π radians), both functions repeat. So the period of both sine and cosine is 2π . Hence, we can find the whole number line wrapped around the unit circle.
Step-by-step explanation:
The correct option is "d".
Given that L = 0.8T²
length of pendulum = 30ft
L= 0.8T²
30 = 0.8T²
T² = 30 / 0.8
T² = 37.5
T = √37.5 = 6.1 seconds
<span>So, 6.1 is the closest to the period in seconds for a pendulum that is 30 ft long.</span>
See the attached figure.
m ∠KAJ = 170° & m ∠LAM = 80°
We should know that :
m ∠KAJ + m ∠LAM + m ∠KAL + m ∠MAJ = 360°
∴ m ∠KAL + m ∠MAJ = 360° - (m ∠KAJ + m ∠LAM)
∴ m ∠KAL + m ∠MAJ = 360° - (170°+80°) = 360° - 250° = 110°
But : m ∠KAL = m ∠MAJ ⇒⇒⇒ <u>Opposite angles.</u>
∴ m ∠MAJ + m ∠MAJ = 110°
∴ 2 * m ∠MAJ = 110°
∴ m ∠MAJ = 110° ÷ 2 = 55°
<u>So, the answer is : m ∠MAJ = 55°</u>
Answer: 0.0035
Step-by-step explanation:
Given : The readings on thermometers are normally distributed with a mean of 0 degrees C and a standard deviation of 1.00 degrees C.
i.e.
and
Let x denotes the readings on thermometers.
Then, the probability that a randomly selected thermometer reads greater than 2.17 will be :_
![P(X>2.7)=1-P(\xleq2.7)\\\\=1-P(\dfrac{x-\mu}{\sigma}\leq\dfrac{2.7-0}{1})\\\\=1-P(z\leq2.7)\ \ [\because\ z=\dfrac{x-\mu}{\sigma}]\\\\=1-0.9965\ \ [\text{By z-table}]\ \\\\=0.0035](https://tex.z-dn.net/?f=P%28X%3E2.7%29%3D1-P%28%5Cxleq2.7%29%5C%5C%5C%5C%3D1-P%28%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5Cleq%5Cdfrac%7B2.7-0%7D%7B1%7D%29%5C%5C%5C%5C%3D1-P%28z%5Cleq2.7%29%5C%20%5C%20%5B%5Cbecause%5C%20z%3D%5Cdfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D%5D%5C%5C%5C%5C%3D1-0.9965%5C%20%5C%20%5B%5Ctext%7BBy%20z-table%7D%5D%5C%20%5C%5C%5C%5C%3D0.0035)
Hence, the probability that a randomly selected thermometer reads greater than 2.17 = 0.0035
The required region is attached below .