NC0 + nC3 + nC6 + nC9 + ... what does this equal to?

Mathematics

1 answer:

galina1969 [7]3 years ago

5 0

NC18 apparently this is too short of an answer and is not explained well

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Simplify -(2x-3)+7-4x

storchak [24]

-6x + 10 is the answer!!!!

3 0

2 years ago

Help please i need answer ASAP

Juli2301 [7.4K]

If the result is reflected over x-axis and vertically stretched by a factor of 5, the result will be g(x) = -5(x - 1)²

<h3>Transformation of functions</h3>

Transformation is a way of changing the position of an object o the xy plane. Given the parent function expressed as

f(x) = x^2

If the function is shifted to the left by unit, then we will have h(x) = (x - 1)²

If the result is reflected over x-axis and vertically stretched by a factor of 5, the result will be g(x) = -5(x - 1)²

Learn more on translation here: brainly.com/question/12861087

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8 0

1 year ago

There are two games involving flipping a coin. In the first game you win a prize if you can throw between 45% and 55% heads. In

Nina [5.8K]

Answer:

d) 300 times for the first game and 30 times for the second

Step-by-step explanation:

We start by noting that the coin is fair and the flip of a coin has a probability of 0.5 of getting heads.

As the coin is flipped more than one time and calculated the proportion, we have to use the <em>sampling distribution of the sampling proportions</em>.

The mean and standard deviation of this sampling distribution is:

$\mu_p=p\\\\ \sigma_p=\sqrt{\dfrac{p(1-p)}{N}}$

We will perform an analyisis for the first game, where we win the game if the proportion is between 45% and 55%.

The probability of getting a proportion within this interval can be calculated as:

$P(0.45$

referring the z values to the z-score of the standard normal distirbution.

We can calculate this values of z as:

$z_H=\dfrac{p_H-\mu_p}{\sigma_p}=\dfrac{(p_H-p)}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_H-p)>0\\\\\\z_L=\dfrac{p_L-\mu_p}{\sigma_p}=\dfrac{p_L-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_L-p)$

If we take into account the z values, we notice that the interval increases with the number of trials, and so does the probability of getting a value within this interval.

With this information, our chances of winning increase with the number of trials. We prefer for this game the option of 300 games.

For the second game, we win if we get a proportion over 80%.

The probability of winning is:

$P(p>0.8)=P(z>z^*)$