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I am Lyosha [343]
3 years ago
9

What type of insurance pays personal expenses due to an accident that was the fault of the other driver who did not have suffici

ent insurance? O A collision O B. gap insurance OC. liability insurance D. uninsured motorist​
Mathematics
1 answer:
Sphinxa [80]3 years ago
5 0

The type of insurance which pays personal expenses due to an accident that was the fault of the other driver who did not have sufficient insurance is:

  • D. uninsured motorist​

<h3>What is Insurance?</h3>

This refers to the coverage or protection which a person has for an unexpected event and pays periodic fee called premium.

With this in mind, if a motorist is involved in an accident, and the accident was the fault of the other driver who is not well insured, then uninsured motorist insurance covers the bill.

Read more about insurance here:

brainly.com/question/25855858

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If the mean heights of three groups of students consisting of 20,16 and 14 students are 167m, 150m and 1.40m respectively find t
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Answer:

1.54 m

Step-by-step explanation:

Given information:

<u>Three groups of students</u>

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\begin{aligned}\implies \sf mean\:height & =\dfrac{(20 \times 1.67)+(16 \times 1.50)+(14 \times 1.40)}{50}\\\\ & = \dfrac{33.4+24+19.6}{50}\\\\ & = \dfrac{77}{50}\\\\ & = 1.54\:\: \sf m\end{aligned}

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2 years ago
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For what values of x does the curve y^2x^3-15x^2=4 have horizontal tangent lines
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\frac{d}{dx}(y^2x^3-15x^2=4) \\ \\&#10;\frac{d}{dx}(y^2x^3)-\frac{d}{dx}(15x^2)=\frac{d}{dx}(4) \\ \\&#10;(y^{2})'x^3+(x^3)'y^2-15(x^2)'=0 \\ \\ 2yy'x^3+3x^2y^2-30x=0 \\ \\&#10;y'=\frac{dy}{dx}=\frac{30x-3x^2y^2}{2x^3y}=\frac{x(30-3xy^2)}{2x^3y} \\ \\&#10;y'=\frac{30-3xy^2}{2x^2y}

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\frac{30-3xy^2}{2x^2y}=0 \\ \\ That \ is, \&#10;when: \\ \\ 30-3xy^2=0 \\ \\ \therefore xy^2=10 \rightarrow y^2=\frac{10}{x}

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\frac{10}{x}x^3-15x^2=4 \\ \\ \therefore&#10;10x^2-15x^2=4 \\ \\ \therefore x^2=-\frac{4}{5}

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Answer:

The range is −5≤y≤−1 - 5 ≤ y ≤ - 1 .

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