You plan to purchase dental insurance for your three remaining years in school. The insurance makes a​ one-time payment of ​
1 answer:
Answer:
(a)D. The value of the payout depends on whether you will need major, minor, or no dental repair over the next 3 years.
(b)Discrete
(c) See below
Step-by-step explanation:
- For a major dental repair, the insurance company pays $1200
- For a minor dental repair, the insurance company pays $160
- For no dental repair, the insurance company pays $0
(a) X= payout of dental insurance a random variable
The payout of dental insurance, X is a random variable because the value of the payout depends on whether you will need major, minor, or no dental repair over the next 3 years. The correct option is D.
(b) X is a discrete variable. This is because its values are whole numbers.
The possible values are $1200, $160 and $0.
(c)
- The probability of requiring a major dental repair (with payout of $1200) is 6%.
- The probability of requiring a minor dental repair (with payout of $160) is 59%.
- The probability of requiring no dental repair (with payout of $0) is 35%.
Therefore, the probability distribution of X is given below:
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Answer:
Step-by-step explanation:
We have the system of equations:
We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:
Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.
In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:
Where the (x, y, z) is our solution set.
Reducing:
With our system, we will have the following matrix:
What we should begin by doing is too see how we can change each row to the reduced-form.
Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:
Now, we can multiply R₂ by -1/5. This yields:
From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:
We can eliminate the -5 in R₃ by adding 5(R₂). This yields:
We can now reduce R₃ by multiply it by -1/4:
Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:
And finally, we can eliminate the second 1 by adding -(R₃):
Therefore, our solution set is
And we're done!