<h3>
Answer: 5</h3>
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Explanation:
Let's consider the expression (x-y)^2. It expands out to x^2-2xy+y^2. The terms are:
Each of those terms either has a single variable with an exponent of 2, or has the exponents add to 2. Think of 2xy as 2x^1y^1.
In short, this means that the degree of each monomial term is 2.
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Now consider (x-y)^3. It expands out into x^3-3x^2y+3xy^2+y^3.
We have terms that either have a single variable and the exponent is 3, or the exponents add to 3. The degree of each term is 3.
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This pattern continues.
In general, for (x-y)^n, where n is any positive whole number, the degree of each term in the expansion is n. If you picked any term, added the exponents, then the exponents will add to n.
Answer:
- P(≥1 working) = 0.9936
- She raises her odds of completing the exam without failure by a factor of 13.5, from 11.5 : 1 to 155.25 : 1.
Step-by-step explanation:
1. Assuming the failure is in the calculator, not the operator, and the failures are independent, the probability of finishing with at least one working calculator is the complement of the probability that both will fail. That is ...
... P(≥1 working) = 1 - P(both fail) = 1 - P(fail)² = 1 - (1 - 0.92)² = 0.9936
2. The odds in favor of finishing an exam starting with only one calculator are 0.92 : 0.08 = 11.5 : 1.
If two calculators are brought to the exam, the odds in favor of at least one working calculator are 0.9936 : 0.0064 = 155.25 : 1.
This odds ratio is 155.25/11.5 = 13.5 times as good as the odds with only one calculator.
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My assessment is that there is significant gain from bringing a backup. (Personally, I might investigate why the probability of failure is so high. I have not had such bad luck with calculators, which makes me wonder if operator error is involved.)
As A Simplified Fraction She Only Has 1/4 Of Her Allowance Left
Answer:
2
Step-by-step explanation:
i got 20/8 from multiplying same with 1/2
20/8 - 1/2 = 2
