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3 years ago

In a chess club the probability that Shaun will beat Mike is 3/8 .

Mathematics

1 answer:

romanna [79]3 years ago

3 0

Answer:

$\frac{10}{56} = 0.1786$ probability that Shaun loses both games

Step-by-step explanation:

Games are independent, so we find each separate probability, and multiply them.

In a chess club the probability that Shaun will beat Mike is 3/8.

So $1 - \frac{3}{8} = \frac{8}{8} - \frac{3}{8} = \frac{5}{8}$ probability that Shaun loses.

The probability that Shaun will beat Tim is 5/7 .

So $1 - \frac{5}{7} = \frac{2}{7}$ probability that Shaun loses.

What is the probability that Shaun loses both games?

This is:

$p = \frac{5}{8} \times \frac{2}{7} = \frac{5*2}{8*7} = \frac{10}{56} = 0.1786$

$\frac{10}{56} = 0.1786$ probability that Shaun loses both games

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Volume of the cone = 130.83 cubic units

Step-by-step explanation:

The question us incomplete without the diagram of the cone and it's dimensions as this would enable us get the exact answer.

Find attached the diagram used in solving the solution.

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3 years ago

If a polynomial function f(x) has roots -8, 1, and 6i, what must also be a root of f(x)?

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Answer:

it must also have the root : - 6i

Step-by-step explanation:

If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.

This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:

$(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2$

where the imaginary unit has disappeared, making the expression real.

So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)

Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.

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