5. When I sine an angle like 60°, it gives me a value like 0.866025403. What does this number represent?

Mathematics

1 answer:

PolarNik [594]3 years ago

5 0

Answer:

See below

Step-by-step explanation:

The number you described is the same as $sin(60^{\circ})=\frac{\sqrt{3}}{2}$ where the sine of an angle is the ratio between a right triangle's opposite side to the angle and the hypotenuse. So, in this case, if we had a right triangle with a height of $\sqrt{3}$ units and a hypotenuse of 2 units, the ratio between the two sides will result in the value you provided. This right triangle in particular would be a 30-60-90 triangle.

In the case of a unit circle, it’s the y-coordinate of the point where a 60° angle in the standard position intersects a unit circle and a right triangle is created from that.

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Two cars start from the same point driving in opposite directions. Both cars drive 6miles. One car stops, the other car makes a

dimaraw [331]

Answer:

<u>14.4 mile</u>²

Explanation:

1. By driving 6 miles each, the cars will be 6 miles + 6 miles = 12 miles appart.

2. When one car stops and the other car makes a 90° left-hand turn and drive 8 miles, you can model the situation with a right triangle, where one leg is 12 miles, the other leg is 8 miles, and the hypotenuse is the distance that separates the cars.

Hence, you must use Pythagoras to find that distance.

3. Pythagoras

hypotenuse² = leg₁² + leg₂²

hypotenuse = (12 mile)² + (8 mile)² = 144 mile² + 64 mile² = 208 mile²

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8 0

3 years ago

A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses

Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = average age of the random sample of horses with colic = 12 yrs

$\mu$ = average age of all horses seen at the veterinary clinic = 10 yrs

$\sigma$ = standard deviation of all horses coming to the veterinary clinic = 8 yrs

n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P( $\bar X$ $\geq$ 12)

P( $\bar X$ $\geq$ 12) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{12-10}{\frac{8}{\sqrt{60} } }$ ) = P(Z $\geq$ 1.94) = 1 - P(Z < 1.94)