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3 years ago

How much interest is earned in 4 years on $5,000 at a simple interest rate of 3%?

Mathematics

1 answer:

Sever21 [200]3 years ago

5 0

Answer:

the answer is 600

Step-by-step explanation:

$5,000×3%×4years÷100%=$600

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3 years ago

The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event th

choli [55]

Answer:

Step-by-step explanation:

From the given information:

The uniform distribution can be represented by:

$f_x(x) = \dfrac{1}{1500} ; o \le x \le \ 1500$

The function of the insurance is:

$I(x) = \left \{ {{0, \ \ \ x \le 250} \atop {x -20 , \ \ \ \ \ 250 \le x \le 1500}} \right.$

Hence, the variance of the insurance can also be an account forum.

$Var [I_{(x}) = E [I^2(x)] - [E(I(x)]^2$

here;

$E[I(x)] = \int f_x(x) I (x) \ sx$

$E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250) \ dx$

$= \dfrac{1}{1500 } \dfrac{(x - 250)^2}{2} \Big |^{1500}_{250}$

$\dfrac{5}{12} \times 1250$

Similarly;

$E[I^2(x)] = \int f_x(x) I^2 (x) \ sx$

$E[I(x)] = \dfrac{1}{1500} \int ^{1500}_{250{ (x- 250)^2 \ dx$

$= \dfrac{1}{1500 } \dfrac{(x - 250)^3}{3} \Big |^{1500}_{250}$

$\dfrac{5}{18} \times 1250^2$

∴

$Var {I(x)} = 1250^2 \Big [ \dfrac{5}{18} - \dfrac{25}{144}]$

Finally, the standard deviation of the insurance payment is:

$= \sqrt{Var(I(x))}$

$= 1250 \sqrt{\dfrac{5}{48}}$

≅ 404

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3 years ago