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andriy [413]
3 years ago
15

Jeanine paid an annual premium of $1,200 in total liability coverage for her car, including up to $400,000 in bodily injury cove

rage and $100,000 in property damage coverage. Twelve years into her policy, Jeanine caused an accident that resulted in the other driver claiming $40,000 in medical costs and $15,000 in car damage. Jeanine's insurance company paid the claim. Did the cost of the benefit of transferring the risk to the insurance company outweigh the cost of the premium? a.Yes, the cost of the annual premium for 12 years was more than the accident claims b.Yes, the cost of the annual premium for 12 years was less than the accident claims c.No, the cost of the annual premium for 12 years was the same as the accident claims d.No, the cost of the annual premium for 12 years was more than the accident claims
Mathematics
1 answer:
Vinil7 [7]3 years ago
3 0

The answer is B- Yes, the cost of the annual premium for 12 years was less than the accident claims.

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Please help with this question ​
zalisa [80]

Answer:

4cm^2

Step-by-step explanation:

First we need to find the length of the height of the triangle using Pythagoras:

a^2+b^2=c^2 (Substitute in what we have)

a^2+3√2(^2)=2√5(^2)   =   a^2 + 18 = 20 -> 20-18=2 √2=BD

Now we can find the area:

AD+DC= 3√2 + √2 = 4√2 x √2 = 8/2 = 4

6 0
3 years ago
Help fast please!!! due soon
aliina [53]

Answer           2.8  

Step-by-step explanation:

yeah

4 0
3 years ago
ASAP PLEASE!! The morning temperature on Saturday was 21.5°. By 10:00 it had warmed up 6.25°. What is the new temperature?
zaharov [31]

The temperature would be (21.5 + 6.25) which is 27.75


4 0
3 years ago
The question is Find 6/7 of 1 2/7
alekssr [168]
1 2/7 = 9/7
9/7 * 6/7 = 54/49
3 0
3 years ago
If f(x) = 3x - 1 and 2f(b) = 28, what is the value of f(2b)?
Vlad1618 [11]
<h3>Answer:  29</h3>

=================================

Work Shown:

Solve the second equation for f(b)

2f(b) = 28

f(b) = 28/2

f(b) = 14

----------

Plug in x = b and solve for b

f(x) = 3x-1

f(b) = 3b-1 ... replace every x with b

14 = 3b-1 ..... plug in f(b) = 14

3b-1 = 14

3b = 14+1

3b = 15

b = 15/3

b = 5

-----------

f(x) = 3x-1

f(2b) = 3(2b)-1 .... replace every x with 2b

f(2b) = 3(2*5)-1 .... plug in b = 5

f(2b) = 3*10-1

f(2b) = 30-1

f(2b) = 29

3 0
3 years ago
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