A thirty year old women decides to pay $300 for a one-year life insurance policy with coverage of $100,000. The probability of her living through the year is 0.9976. What is her expected value for the insurance policy?

So then the expected value for this case on the year described for the woman is --$59.28, and that means a net loss.

Step-by-step explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

Solution to the problem

For this case the woman needs to pay $300 for a one year life insurance and we know that the probability that she will live the year is 0.9976. So then the probability that she not live on the year is 1-0.9976=0.0024 by the complement rule , and on this case the company needs to pay 100000 for the life insurance.

So then we can use the definition of expected value given by:

And replacing the values given we have:

So then the expected value for this case on the year described for the woman is --$59.28, and that means a net loss.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

you can use a^2+ b^2=c^2