Ask question

Ask question

All categories

3 years ago

9

Identify the equation in slope-intercept form for the line containing the point (0,7) and perpendicular to y=−5/4x+11/4. HELP PL

EASE!!

2 answers:

777dan777 [17]3 years ago

6 0

To write the equation of a line I need slope and y intercept (0,y)

I have y intercept (0,7) of +7 as written in the equation

The equation y=-5/4x + 11/4 has a slope of -5/4

To get perpendicular slope is negated reciprocal

so take -5/4 and flip it -4/5 and change the sign (in this case negative to positive) 4/5

so the new equation is y=4/5x +7 (third one)

Nookie1986 [14]3 years ago

3 0

Answer:

To write the equation of a line I need slope and y intercept (0,y)I have y intercept (0,7) of +7 as written in the equationThe equation y=-5/4x + 11/4 has a slope of -5/4To get perpendicular slope is negated reciprocalso take -5/4 and flip it -4/5 and change the sign (in this case negative to positive) 4/5so the new equation is y=4/5x +7 (third one)

You might be interested in

State whether the lines are parallel, perpendicular, coinciding, or neither: 6x+3y=–15 and y–3=–2x

lianna [129]

Answer:

Parallel

<u>Step-By-Step Explanation:</u>

Put the Function in Slope Intercept Form and Find the Slope of 6x+3y = 15

6x+3y = 15

3y = -6x + 15

3y/3 = -6x/3 + 15/3

y = -2x + 5

<u>We can see that the slope of 6x+3y = 15 is -2</u>

Put the Function in Slope Intercept Form and Find the Slope of y–3=–2x

y–3=–2x

y = -2x + 3

Here are our two Functions In Slope Intercept Form

y = -2x + 5

y = -2x + 3

<u>Remember the m = slope and the b = y-intercept</u>

y = mx + b

y = -2x + 5

y = -2x + 3

------------------------------------------------------------------------------------------------------

We can see both equations have the same slope of -2 so this means they could be parallel because parallel functions have the same slope but coinciding functions have the same slope too. To tell if the two functions are coinciding, the functions need to have the same slope and the same y-intercept. Looking at the two functions, we can see they have the same slope of -2 but their y-intercept are different so this makes the two functions parallel.

4 0

3 years ago

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

Determine whether the value is from a discrete or continuous data set. Number of coins in a jar is 78 number of coins in a jar i

Phoenix [80]

Determine whether the value is from a discrete or continuous data set. Number of coins in a jar is 78 number of coins in a jar is 78

Answer: Number of coins in a jar is from a discrete data set. Because the given variable is countable in a finite amount of time.

If a variable can take on any value between two specified values, it is called a continuous variable; otherwise, it is called a discrete variable.

3 0

3 years ago

Help needed due today

kobusy [5.1K]

Answer:

1 1/4 inches deeper

Step-by-step explanation:

3 0

2 years ago

What is 2 thirds of 60

Mkey [24]

You have to set up a proportion so 2 over 3= x over 60. So multiply 2 times 60= 120 divided by 3 which is 40

6 0

3 years ago