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mafiozo [28]
4 years ago
7

Jake has $25.000 worth of property damage insurance and $1,000- deductible collision insuranceHe caused an accident that damaged

a 2,000 sign, and he also did $ 2,400 worth of damage to another . His car had 2.980 worth of damage done?
a. How much will the insurance company pay under Jake's property damage insurance?

B.How much will the insurance company pay under Jake's collision insurance?

C.How much of the damage must Jake pay for?
Mathematics
1 answer:
Flauer [41]4 years ago
5 0

Answer:

I think b will be good for you

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The answer is answer choice 3
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3 years ago
Arderi Air Conditioning sells one primary model of air conditioner. Each unit of this model costs them $61.25 to produce, but it
svetoff [14.1K]
THE CORRECT ANSWER IS:

Mary sells 43 units; Jay sells 51 units; Abby sells 54 units



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3 years ago
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Can someone pls help for brainlest
uranmaximum [27]

Answer:

a = 6

Step-by-step explanation:

Area = 36 in^{2}

Only one number will make this missing length true.

Obviously it's a square so one number x one number will equal to 36 in^{2}

6 x 6 = 36 in^{2}

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3 years ago
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What is the first step in evaluating the expression below?
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Answer:

brackets, then Parenthesis then the FIRST Multiplication then the division and then math and sub and so on a so for

Step-by-step explanation:

6 0
3 years ago
Sherman has two sequences . The first sequence is described by the explicit rule f(n) = 15n + 4 and the second sequence is descr
Viktor [21]

Answer:

Step-by-step explanation:

Given the explicit function as

f(n) = 15n+4

The first term of the sequence is at when n= 1

f(1) = 15(1)+4

f(1) = 19

a = 19

Common difference d = f(2)-f(1)

f(2) = 15(2)+4

f(2) = 34

d = 34-19

d = 15

Sum of nth term of an AP = n/2{2a+(n-1)d}

S20 = 20/2{2(19)+(20-1)15)

S20 = 10(38+19(15))

S20 = 10(38+285)

S20 = 10(323)

S20 = 3230.

Sum of the 20th term is 3230

For the explicit function

f(n) = 4n+15

f(1) = 4(1)+15

f(1) = 19

a = 19

Common difference d = f(2)-f(1)

f(2) = 4(2)+15

f(2) = 23

d = 23-19

d = 4

Sum of nth term of an AP = n/2{2a+(n-1)d}

S20 = 20/2{2(19)+(20-1)4)

S20 = 10(38+19(4))

S20 = 10(38+76)

S20 = 10(114)

S20 = 1140

Sum of the 20th terms is 1140

7 0
3 years ago
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