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3 years ago

8(12+a) I'm Having trouble with this

Mathematics

2 answers:

Otrada [13]3 years ago

7 0

Answer:

8a+96 or a+12

Step-by-step explanation:

Distribute

KonstantinChe [14]3 years ago

5 0

Answer:

8a+96

Step-by-step explanation:

8(12+a)

= 96+8a

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A formula for finding the area of a rectangle is A = l • w. If you know the area (A) and width (w) of a rectangle, which formula

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Take the area and divide out what you know, so if you know (w) then divide (w) out from the area

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Can the product of 4 and any number always be written as a sum of two equal addends

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Answer: Yes

Step-by-step explanation:

Example : 4 x 2 = 8

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Water freezes at 0 degrees Celsius. A student discovered that adding half a cup of salt to a gallon of water lowers its freezing

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Answer:

-7°C

Step-by-step explanation:

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Freezing point of salt water :

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3 years ago

Only 6 I need need serious help

k0ka [10]

Answer:

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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 <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:

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

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.

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3 years ago