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2 years ago

If f(x)= x^3 - 2x^2, find f(-2)

Mathematics

1 answer:

KATRIN_1 [288]2 years ago

6 0

hope it helps

plz mark brainliest

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Simplify the expression to a + bi form: (-6 - 9i)(-11 + 7i)

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129+57i

Step-by-step explanation:

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G(-3)= 2(-3)-22=-28

This is the correct answer

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2 years ago

The volume of this prism is 4375cm3. The area of the cross-section is 125cm2. Work out x

fomenos

Answer:

x is 35 cm

Step-by-step explanation:

To find the value of x, we will divide the volume by the cross-section

Mathematically, that will be;

4375/125 = 35 cm

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2 years ago

Which of the following functions is graphed below

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D is the answer that is graphed.

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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 sampling distribution of the sampling proportions.

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:

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

The z value is calculated as before:

$z^*=\dfrac{p^*-\mu_p}{\sigma_p}=\dfrac{p^*-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p^*-p)>0$

As (p*-p)=0.8-0.5=0.3>0, the value z* increase with the number of trials (N).

If our chances of winnings depend on P(z>z*), they become lower as z* increases.

Then, we can conclude that our chances of winning decrease with the increase of the number of trials.

We prefer the option of 30 trials for this game.

8 0

2 years ago