All categories

2 years ago

4 questions! Write an equation in slope-intercept form for each of the following:

1 answer:

3 0

1) y= -4x+c

2) y= 3x+c replace x and y to find c , 5=3(2)+c
C= -1
Y=3x-1

3) find slope first m=y2-y1/x2-x1 = 0-(-1)/-2-1 = 1/-3 and now you have y= -1/3x+c to find c just replace any of the given points , 0= -1/3(-2)+c
C= -2/3 and so y= -1/3x-2/3

4) DO THE SAME STEP AS NUMBER 3 and enjoy!

You might be interested in

You have a large jar that initially contains 30 red marbles and 20 blue marbles. We also have a large supply of extra marbles of

Dima020 [189]

Answer:

There is a 57.68% probability that this last marble is red.

There is a 20.78% probability that we actually drew the same marble all four times.

Step-by-step explanation:

Initially, there are 50 marbles, of which:

30 are red

20 are blue

Any time a red marble is drawn:

The marble is placed back, and another three red marbles are added

Any time a blue marble is drawn

The marble is placed back, and another five blue marbles are added.

The first three marbles can have the following combinations:

R - R - R

R - R - B

R - B - R

R - B - B

B - R - R

B - R - B

B - B - R

B - B - B

Now, for each case, we have to find the probability that the last marble is red. So

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8}$

$P_{1}$ is the probability that we go R - R - R - R

There are 50 marbles, of which 30 are red. So, the probability of the first marble sorted being red is $\frac{30}{50} = \frac{3}{5}$ .

Now the red marble is returned to the bag, and another 3 red marbles are added.

Now there are 53 marbles, of which 33 are red. So, when the first marble sorted is red, the probability that the second is also red is $\frac{33}{53}$

Again, the red marble is returned to the bag, and another 3 red marbles are added

Now there are 56 marbles, of which 36 are red. So, in this sequence, the probability of the third marble sorted being red is $\frac{36}{56}$

Again, the red marble sorted is returned, and another 3 are added.

Now there are 59 marbles, of which 39 are red. So, in this sequence, the probability of the fourth marble sorted being red is $\frac{39}{59}$ . So

$P_{1} = \frac{3}{5}*\frac{33}{53}*\frac{36}{56}*\frac{39}{59} = \frac{138996}{875560} = 0.1588$

$P_{2}$ is the probability that we go R - R - B - R

$P_{2} = \frac{3}{5}*\frac{33}{53}*\frac{20}{56}*\frac{36}{61} = \frac{71280}{905240} = 0.0788$

$P_{3}$ is the probability that we go R - B - R - R

$P_{3} = \frac{3}{5}*\frac{20}{53}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{937570} = 0.076$

$P_{4}$ is the probability that we go R - B - B - R

$P_{4} = \frac{3}{5}*\frac{20}{53}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{968310} = 0.0511$

$P_{5}$ is the probability that we go B - R - R - R

$P_{5} = \frac{2}{5}*\frac{30}{55}*\frac{33}{58}*\frac{36}{61} = \frac{71280}{972950} = 0.0733$

$P_{6}$ is the probability that we go B - R - B - R

$P_{6} = \frac{2}{5}*\frac{30}{55}*\frac{25}{58}*\frac{33}{63} = \frac{49500}{1004850} = 0.0493$

$P_{7}$ is the probability that we go B - B - R - R

$P_{7} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{33}{63} = \frac{825}{17325} = 0.0476$

$P_{8}$ is the probability that we go B - B - B - R

$P_{8} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{30}{65} = \frac{750}{17875} = 0.0419$

So, the probability that this last marble is red is:

$P = P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} + P_{7} + P_{8} = 0.1588 + 0.0788 + 0.076 + 0.0511 + 0.0733 + 0.0493 + 0.0476 + 0.0419 = 0.5768$

There is a 57.68% probability that this last marble is red.

What's the probability that we actually drew the same marble all four times?

$P = P_{1} + P_{2}$

$P_{1}$ is the probability that we go R-R-R-R. It is the same $P_{1}$ from the previous item(the last marble being red). So $P_{1} = 0.1588$

$P_{2}$ is the probability that we go B-B-B-B. It is almost the same as $P_{8}$ in the previous exercise. The lone difference is that for the last marble we want it to be blue. There are 65 marbles, 35 of which are blue.

$P_{2} = \frac{2}{5}*\frac{25}{55}*\frac{1}{2}*\frac{35}{65} = \frac{875}{17875} = 0.0490$

$P = P_{1} + P_{2} = 0.1588 + 0.0490 = 0.2078$

There is a 20.78% probability that we actually drew the same marble all four times

3 0

3 years ago

A line plot could be used as a data display for the distribution of test scores in math class. true or false

Andreas93 [3]

A line plot can be used as a data display for the distribution of test scores in math class. Thus, the option to choose is <u>TRUE</u>.

A line plot is a graph that shows data as checkmarks or dots across a number line, indicating the frequency of each value.

The test score in the math test will be discrete data with repeating frequencies. The score of every individual will moreover be a whole number.

The line plot, as we know, marks the dots across the number line, indicating frequency.

Thus, the scores in the math test can be displayed using a line plot, by indicating each type of mark on the line, and dots above those marks for the number of times the same marks occur.

Thus, the option to choose is <u>true</u>.

Learn more about line plots at

brainly.com/question/8989301

#SPJ4

8 0

2 years ago

Find the y-intercept of the line on the graph

nadezda [96]

4 is on the y-intercept. the y axis is vertical and the x axis is horizontal.

4 0

3 years ago

Read 2 more answers