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3 years ago

Given the equation of the line (y-2)=3(x+1) what is the slope

Mathematics

1 answer:

DIA [1.3K]3 years ago

4 0

The slope is 3. If an equation is in the form
Y-Y1=M(X-X1), then M is the slope. therefore, 3 is the slope

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. (6.03) Which of the following best describes the expression 6(y + 3)? (4 points) The product of two constant factors six and t

kolezko [41]

C. The product of a constant factor of six and a factor with the sum of two terms. In this case, 6 is constant, and the sum of two terms is: y+3. Since they are products, you multiple them. Therefore C would be correct.

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3 years ago

What is the percent change from 400 to 500?

faust18 [17]

Answer:

25%

Step-by-step explanation:

8 0

1 year ago

Put me out of my misery! <br> Help me! <br> If possible could you show me how to do this?

prisoha [69]

Answer:

4z+4

Divide each term of 4z+4 by 4 to get 1+z.

1+z

8 0

2 years ago

Read 2 more answers

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.

8 0

3 years ago

Of the 20 students in Emily's class 15 like chocolate cake and the rest like cheesecake how many like cheesecake and what percen

alisha [4.7K]

5 students like cheesecake because if there is 20 students in all and 15 like chocolate and the rest like cheesecake you do 20-15=5

If 5 students like cheesecake out of 20 it would be 5/20 20*5=100 just multiply by 5 5*5=25%

25% of students like cheesecake

4 0

4 years ago

Read 2 more answers