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Marianna [84]
3 years ago
7

. An insurance company offer contracts for a certain peril. In a contract period, which is 6 months, the following payoffs can o

ccur: payoff Prob. Of loss 15000 0.015 12000 0.023 8000 0.028 6000 0.035 Find the average payoff per contract in a contract period. If the company has to cover its operating cost which is a 12% overhead and make a 4% profit what should be the yearly premium which must charge for each contract? What is the probability that a contract will incur losses not exceeding $9,200 during a contract period?

Mathematics
1 answer:
Damm [24]3 years ago
5 0

Answer and Step-by-step explanation:

The answer is attached below

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20 hours put into degrees
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Answer:

  300°

Step-by-step explanation:

If 24 hours is 360°, then ...

  (20 hours)/(24 hours) × 360° = 300°

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Please help I'm stuck on these 2 problems would help so much!!<br><br> Thanks!
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Please Help Me!
alex41 [277]

Answer:

✔️2 sets of corresponding angles

<D and <S

<R and <L

✔️2 sets of corresponding sides

DR and SL

RM and LT

Step-by-step explanation:

When two polygons are congruent, it implies that they have the same shape and size. Therefore, their corresponding angles and sides are congruent to each other.

When naming congruent polygons, the arrangement of the vertices are kept in a definite order of arrangement.

Therefore, Given that polygon DRMF is congruent to SLTO, the following angles and sides correspond to each other:

<D corresponds to <S

<R corresponds to <L

<M corresponds to <T

<F corresponds to <O

For the sides, we have:

DR corresponds to SL

RM corresponds to LT

MF corresponds to TO

FD corresponds to OS.

We can select any two out of these sets of corresponding angles and sides as our answer. Thus:

✔️2 sets of corresponding angles

<D and <S

<R and <L

✔️2 sets of corresponding sides

DR and SL

RM and LT

4 0
3 years ago
What two numbers add to 8 and multiply to 28
anygoal [31]
Find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula

x+y=8
xy=28

x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0

if you have
ax^2+bx+c=0 then x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28

x=\frac{-(-8)+/- \sqrt{(-8)^{2}-4(1)(28)} }{2(1)}
x=\frac{8+/- \sqrt{-48} }{2}
x=\frac{8+/- 4 \sqrt{-3} }{2}
x=4+/- 2 \sqrt{-3}
x=4+/- 2i \sqrt{3}
there are no real numbers that satisfy this
5 0
3 years ago
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