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Yuki888 [10]
2 years ago
12

The policyholder is really tight on cash and wants to trim down their insurance premiums so they can devote more of their budget

to paying off a large credit card debt. Which coverage makes the most sense to eliminate?
A.) Liability coverage, because it is the most expensive part of their bill
B.) Collision coverage, because their car is quite old and the deductible is relatively high
C.) Uninsured motor vehicle coverage, because they already have car insurance, so they won’t ever be uninsured
D.) They cannot eliminate any of this coverage -- it is all required by Federal law
Mathematics
1 answer:
harina [27]2 years ago
7 0

Answer:

Considering the available options, the coverage that makes the most sense to eliminate is the "<u>Collision</u><u> coverage, because their car is quite old and the </u><u>deductible</u><u> is relatively high."</u>

What is Insurance Premium?

The insurance premium is the money by an individual or company to insure some specific properties. Insurance premium usually covers healthcare, auto, home, and life insurance.

However, one of the ways to reduce the mount paid on insurance premiums is to <u><em>reduce coverage on older cars.</em></u>

It is believed that if the car is worth less than 10 times the premium, paying for the coverage is not cost-effective.

Hence, in this case, it is concluded that the correct answer is option A. <u>"</u><u>Collision </u><u>coverage because their car is quite old and the </u><u>deductible </u><u>is relatively </u><u>high</u><u>."</u>

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Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 students. One
Naily [24]

Answer:

The expected value of X is E(X)=\frac{2754}{73} \approx 37.73 and the variance of X is Var(X)=\frac{226192}{5329} \approx 42.45

The expected value of Y is E(Y)=\frac{73}{2} \approx 36.5 and the  variance of Y is Var(Y)=\frac{179}{4} \approx 44.75

Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and  probability mass function p(x). The expected value, denoted by E(X) or \mu_x, is

E(X)=\sum_{x\in D} x\cdot p(x)

The probability mass function p_{X}(x) of X is given by

p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function p_{Y}(x) of Y is given by

p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}

The expected value of X is

E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)

E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73

The expected value of Y is

E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)

E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2

The variance of X is

E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)

E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}

Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45

The variance of Y is

E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)

E(Y^2)=28^2\cdot \frac{1}{4}+32^2\cdot \frac{1}{4} +42^2\cdot \frac{1}{4} +44^2 \cdot \frac{1}{4}\\\\E(Y^2)=196+256+441+484\\\\E(Y^2)=1377

Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75

8 0
3 years ago
One 16 inch pizza serves 4 students how many pizzas are needed for 60 students?​
Salsk061 [2.6K]

Answer:

15 pizzas

Step-by-step explanation:

1 pizza serves 4 students

60 students divided by 4 students (which is 1 pizza)

60/4=15

6 0
3 years ago
About 50 percent of the math questions are multiple choice and 50 percent are grid-in.
OleMash [197]
The answer I think is false
6 0
3 years ago
Read 2 more answers
Need answer ASAP thanks!
barxatty [35]

Answer:

assuming its an annual interest

Okay so 6 percent interest, the bank is paying you.
So with this it’s 6 percent of 1500 and add it to 1500.

You can always find 6 percent of 1500 and then add but here’s a short cut.
Your principle (beginning) balance is 1500.
That’s already 100 percent since thats yoru original value.
You then get added 6 percent interest.
We are jsut adding 6 percent to 100 percent so 106 percent.
Now we solve normally and you’d get the answer faster.

106 percent is 106/100 or 1 3/5 or 1.06

now we multiply
1500 * 1.06 = 1590

Your final balance would be 1590 after the 6 percent interest is added.

5 0
2 years ago
Plzzzzz help I need ​
Lera25 [3.4K]

Answer:

i dont know tbh

Step-by-step explanation:

good luck

8 0
2 years ago
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