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2 years ago

Give the values of a, b, and c needed to write the equation's general form.

1 answer:

5 0

The values of a, b, and c are needed to write the equation's general form will be a = 1; b = -3; c = 0.Option A is corect.

<h3>What is the equation?</h3>

A mathematical statement consisting of an equal symbol between two algebraic expressions with the same value is known as an equation.

Given equation;

x²-3x=0

The quadratic equation in the form is;

ax² + bx + c = 0

On putting the values of the options one by one we will get the answer;

1 x²+(-3)x+0

x²-3x=0

Hence, option A is corect.

To learn more, about equations, refer;

brainly.com/question/10413253

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What is the slope of this graph? −14 4 14 −4 Number graph ranging from negative 10 to 10 on both the x and y axis. A line passes

spin [16.1K]

The slope of this graph is - 4

Step-by-step explanation:

The slope of a line is $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ , where

$(x_{1},y_{1})$ and $(x_{2},y_{2})$ are two points on the line
The slope of a horizontal line is zero
The slope of a vertical line is undefined

∵ The graph is a line

∵ The line passes through points (1 , -2) and (0 , 2)

∴ $x_{1}$ = 1 and $x_{2}$ = 0

∴ $y_{1}$ = -2 and $y_{2}$ = 2

- Substitute these values in the rule below

∵ $m=\frac{y_{2}-y_{1}}{x_{2}-x_{2}}$

∴ $m=\frac{2--2}{0-1}=\frac{4}{-1}$

∴ m = - 4

The slope of this graph is - 4

Learn more:

You can learn more about the slope of a line in brainly.com/question/4152194

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7 0

3 years ago

Read 2 more answers

Which option below best describes the polynomial 10x^3 ?

MAXImum [283]

The answer is C, there’s only 1 term= monomial

5 0

3 years ago

Lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. a bank conducts inter

Otrada [13]

Part A:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector correctly determined that a selected person is saying the truth has a probability of 0.85
Thus p = 0.85

Thus, the probability that the lie detector will conclude that all 15 are telling the truth if all 15 applicants tell the truth is given by:

 $P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(15)={ ^{15}C_{15}(0.85)^{15}(0.15)^0} \\ \\ =1\times0.0874\times1=0.0874$


Part B:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.25
Thus p = 0.15

Thus, the probability that the lie detector will conclude that at least 1 is lying if all 15 applicants tell the truth is given by:

$P(X)={ ^nC_xp^xq^{n-x}} \\ \\ \Rightarrow P(X\geq1)=1-P(0) \\ \\ =1-{ ^{15}C_0(0.15)^0(0.85)^{15}} \\ \\ =1-1\times1\times0.0874=1-0.0874 \\ \\ =0.9126$

Part C:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The mean is given by:

$\mu=npq \\ \\ =15\times0.15\times0.85 \\ \\ =1.9125$

Part D:

Given that lie detectors have a 15% chance of concluding that a person is lying even when they are telling the truth. Thus, lie detectors have a 85% chance of concluding that a person is telling the truth when they are indeed telling the truth.

The case that the lie detector wrongly determined that a selected person is lying when the person is actually saying the truth has a probability of 0.15
Thus p = 0.15

The probability that the number of truthful applicants classified as liars is greater than the mean is given by:

 $P(X\ \textgreater \ \mu)=P(X\ \textgreater \ 1.9125) \\ \\ 1-[P(0)+P(1)]$

 $P(1)={ ^{15}C_1(0.15)^1(0.85)^{14}} \\ \\ =15\times0.15\times0.1028=0.2312$

8 0

4 years ago

Please Help with this Problem!

tatiyna

Answer: 3.5, 4.5, 9.5, 3.5

Step-by-step explanation:

Look at the image below to see where A, B, C, and D are.

A + B = 8

B + D = 8

A + C = 13

C - D = 6

we can see that A + B = 8 and D + B = 8, so A = D

substitute this into A + C = 13 to get D + C = 13

from D + C = 13 we can get D = 13 - C

plug this into C - D = 6 to get C - (13 - C) = 6

2C - 13 = 6

2C = 19

C = 9.5

Now we can find D = 13 - C = 13 - 9.5 = 3.5

D = 3.5

Now we can find A = D = 3.5

A = 3.5

Now we can find B from A + B = 8

B = 8 - A = 8 - 4.5 = 4.5

B = 4.5

6 0

3 years ago

The distribution of the amount of money spent by students for textbooks in a semester is approximately normal in shape with a me

nadezda [96]

The answer is between $175 and $295. The standard deviation rule tells us that for distributions that have the normal shape, approximately 99.7% of the observations fall within 3 standard deviations of the mean. Indeed, 175 = 235 − 3 * 20 and 295 = 235 + 3 * 20 are exactly 3 standard deviations below and above the mean, respectively.

4 0

4 years ago