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3 years ago

WILL GIVE BRAINLIEST PLEASE HELP

Mathematics

1 answer:

Goryan [66]3 years ago

8 0

Answer:

d. $1617

Step-by-step explanation:

The problem parameters are used to identify the row and column to use in each table. The values at that location are added and multiplied by the driver rating factor to get the annual premium. (We aren't exactly sure whether we're computing Ivan's premium or Miller's premium, as the problem text seems ambiguous on that point.)

The (raw) cost of liability insurance is at the intersection of the column for 50/100 bodily injury and the row for 50,000 property damage. It is $344.

The (raw) cost of collision coverage is at the intersection of the column for rating group 13 and the row for age group B. It is $284.

The (raw) cost of comprehensive coverage is at the intersection of the column for rating group 13 and the row for age group B. It is $107.

To get the annual premium the sum of these table values is multiplied by the driver rating factor of 2.2:

annual premium = 2.2×($344 +284 +107) = 2.2×$735 = $1617

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Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

$AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta$ (1)

Where:

$AC$ - The distance from A to C, measured in miles.

$AB$ - The distance from A to B, measured in miles.

$BC$ - The distance from B to C, measured in miles.

$\theta$ - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

$AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta$

$\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}$

$\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right)$ (2)

If we know that $BC = 7\,mi$ , $AB = 8\,mi$ , $AC = 5\,mi$ , then the bearing from island B to island C:

$\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]$

$\theta \approx 38.213^{\circ}$

The bearing needed to navigate from island B to island C is approximately 38.213º.

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3 years ago

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Sorry that's all i know

<h3>that is in my answer</h3>

<h2>#carryonlearning:-D</h2>

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