The values of events A, B, and C are provided. Compare the probabilityof event A occurring, given that event C occurred to the probability ofEvent B happening, given that event C occurred (Compare P(A/C) toP(BIC)] Which event is more likely?P(A) = 0.45P(B) = 0.30P(C) = 0.25

Answers

Answer 1

The conditional probability P(A/C) is given by

[tex]P(A|C)=\frac{P(A\cap C)}{P(C)}[/tex]

If event A is independent to event C, we can write

[tex]P(A|C)=\frac{P(A)\cdot P(C)}{P(C)}=P(A)[/tex]

Similary, if event B is independent to event C, we get

[tex]P(B|C)=\frac{P(B)\cdot P(C)}{P(C)}=P(B)[/tex]

Then, by comparing both results we can see that event A is more likey than event B.


Related Questions

A(0,3) B(1,6) C(4,6) D(5,3) rotate it around the origin 270 degrees clockwise
Please answer with the coordinates.

Answers

The rule for a rotation by 270° about the origin is (x,y)→(y,−x)

so i guess A?

A circular arc has measure of 4 cm and is intercepted by a central angle of 73°. Find the radius r of the circle. Do not round any intermediate computations, and round your answer to the nearest tenth.r= __ cm

Answers

The arc lenghr is given by:

[tex]s=r\theta[/tex]

where s is the arc lenght, r is tha raidus and theta is the angle measure in radians. Since in our problem the angle is given in degrees we have to convert it to radians, to do this we have to multiply the angle by the factor:

[tex]\frac{\pi}{180}[/tex]

Then:

[tex]\theta=(73)(\frac{\pi}{180})[/tex]

Plugging the value of the arc lenght and the angle in the first formula, and solving for r we have:

[tex]\begin{gathered} 4=r(73)(\frac{\pi}{180}) \\ r=\frac{4\cdot180}{73\cdot\pi} \\ r=3.1 \end{gathered}[/tex]

Therefore, the radius of the circle is 3.1 cm.

If the two expressions are equivalent, find value of x

Answers

[tex]\frac{5}{x}-\frac{1}{3}=\frac{1}{x}[/tex]

1. Subtract 1/x in both sides of the equation:

[tex]\begin{gathered} \frac{5}{x}-\frac{1}{x}-\frac{1}{3}=\frac{1}{x}-\frac{1}{x} \\ \\ \frac{4}{x}-\frac{1}{3}=0 \end{gathered}[/tex]

2. Add 1/3 in both sides of the equation:

[tex]\begin{gathered} \frac{4}{x}-\frac{1}{3}+\frac{1}{3}=0+\frac{1}{3} \\ \\ \frac{4}{x}=\frac{1}{3} \end{gathered}[/tex]

3. Multiply both sides of the equation by x:

[tex]\begin{gathered} x\cdot\frac{4}{x}=x\cdot\frac{1}{3} \\ \\ 4=\frac{x}{3} \end{gathered}[/tex]

4. Multiply both sides of the equation by 3:

[tex]\begin{gathered} 4\cdot3=\frac{x}{3}\cdot3 \\ \\ 12=x \\ \\ \text{ Rewrite} \\ x=12 \end{gathered}[/tex]Then, the value of x is 12

Translate into proportion 16.4 is 45% of what number ?

Answers

Given:

16.4 is 45%

[tex]\begin{gathered} 16.4\times\frac{45}{100}=\frac{b}{1} \\ \frac{16.4}{b}=\frac{100}{45} \end{gathered}[/tex]

Hence, the required option is D.

If Martha is x years old and her mother is 5 times older and the sum of their age is 88, how old is Martha?

Answers

Let Martha age = x years old

Since her mother's age is 5 times her age, then

Her mother = 5(x) = 5x years old

Since the sum of their ages is 84, then

Add x and 5x, then equate the sum by 84

[tex]\begin{gathered} x+5x=84 \\ 6x=84 \end{gathered}[/tex]

Divide both sides by 6 to find x

[tex]\begin{gathered} \frac{6x}{6}=\frac{84}{6} \\ x=14 \end{gathered}[/tex]

Martha is 14 years old

In △GHI, m∠G = (9x - 2), m∠H = (3x - 19), and m∠I = (3x + 6)". Find m∠G.

Answers

Answer:

m∠G = 115 degrees

Explanation:

The sum of the angles in a triangle is 180 degrees.

In triangle GHI:

[tex]\begin{gathered} m\angle G+m\angle H+m\angle I=180\degree \\ \implies9x-2+3x-19+3x+6=180\degree \end{gathered}[/tex]

First, solve for x:

[tex]\begin{gathered} 9x+3x+3x-2-19+6=180\degree \\ 15x-15=180\degree \\ 15x=180+15 \\ 15x=195 \\ x=\frac{195}{15} \\ x=13 \end{gathered}[/tex]

Therefore, the measure of angle G is:

[tex]\begin{gathered} m\angle G=9x-2 \\ =9(13)-2 \\ =117-2 \\ m\angle G=115\degree \end{gathered}[/tex]

The measure of angle G is 115 degrees.

I will give brainliest if you help me with this problem not joking

Answers

Answer: 9+6+-6+-7

Step-by-step explanation:

im not sure thats my guess tho

the ratio of isabella's money to Shane's money is 5:10.if Isabelle has $55 how much money do Shane have?what about they have together?

Answers

[tex]\begin{gathered} \text{ratio 5:10} \\ \text{isabelle has 55 then} \\ \frac{55}{5}=11 \end{gathered}[/tex][tex]\begin{gathered} In\text{ order to conserve the ratio, it must ocurr that} \\ \text{Shane has 10}\cdot11=110 \\ \text{why? because} \\ \frac{110}{10}=\frac{55}{5}=11 \end{gathered}[/tex][tex]\text{Together Isabelle and Shane have 110+55=165}[/tex]

turn the expression from radical form to exponential expression in fractional form. No need to evaluate just be out in simplest form

Answers

To answer this question, we need to remember the next property of radicals:

[tex]\sqrt[n]{a^m}=a^{\frac{m}{n}}[/tex]

In this case, we have that:

[tex]\sqrt[3x]{5}[/tex]

And we can see that the exponent for 5 is m = 1. Therefore, we can rewrite the expression as follows:

[tex]\begin{gathered} \sqrt[3x]{5}=5^{\frac{1}{3x}} \\ \end{gathered}[/tex]

In summary, therefore, we can say that the radical form to an exponential in fractional form is:

[tex]undefined[/tex]

Find the restricted values of x for the following rational expression. If there are no restricted values of x,indicate "No Restrictions".x² +8x² - x - 12AnswerHow to enter your answer (opens in new window)Separate multiple answers with commas.KeypadKeyboard ShortcutsSelecting a radio button will replace the entered answer value(s) with the radio button value. If the radiobutton is not selected, the entered answer is used.

Answers

Answer:

To find the restricted values of x for the given rational expression,

[tex]\frac{x^2+8}{x^2-x-12}[/tex]

The above expression is defined only when x^2-x-12 not equal to 0.

x values are restricted for the solution of x^2-x-12=0

To find the values of x when x^2-x-12=0.

Consider, x^2-x-12=0

we get,

[tex]x^2-x-12=0[/tex][tex]x^2-4x+3x-12=0[/tex][tex]x\left(x-4\right)+3\left(x-4\right)=0[/tex]

Taking x-4 as common we get,

[tex]\left(x-4\right)\left(x+3\right)=0[/tex]

we get, x=4,x=-3

The restricted values of x are 4,-3.

we get,

[tex]x\ne4,-3[/tex]

Answer is:

[tex]x\ne4,-3[/tex]

4. Find the midpoint of DK, given the coordinates D (-10, -4) and K is located at the origin.m:|| m:1 m:Midpoint:Equation of the line:

Answers

The midpoint between two coordinates can be calculated using the equation

[tex]m=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]

Point D has the coordinates (-10, -4). The problem stated that point K is located at the origin, hence, we can say that its coordinates are (0, 0).

Using the formula stated above to solve the coordinates of the midpoint, we get

[tex]\begin{gathered} m=(\frac{-10+0}{2},\frac{-4+0}{2}) \\ m=(\frac{-10}{2},-\frac{4}{2}) \\ m=(-5,-2) \end{gathered}[/tex]

Answer: The midpoint of the line segment DK is located at (-5,-2).

Which transformations can be used to carry ABCD ontoitself? The point of rotation is (3, 2). Check all that apply.yB1 23 4 5 6A. Translation four units to the rightB. Dilation by a factor of 2C. Rotation of 180°D. Reflection across the line x = 3

Answers

Answer:

C. Rotation of 180°

D. Reflection across the line x = 3

Explanation:

The point of rotation = (3, 2)

From the diagram:

The center of the rectangle = (3, 2)

Note that:

The point of rotation = The center of the rectangle

Therefore, the rectangle ABCD will be reflected across its center, x =3 and y = 2

Also note that ABCD has two lines of symmetry. Therefore, a rotation by 180 degrees will carry ABCD back to itself

Does the point (2, 6) satisfy the inequality 2x + 2y ≥ 16?
yes
no

Answers

No the answer is no because 4+21 =21 yes

Solve for y and show steps 75-3.5y-4y=4y+6

Answers

Solve for y;

[tex]\begin{gathered} 75-3.5y-4y=4y+6 \\ \text{Collect like terms, which means the values with y would be moved to one side} \\ \text{And the values without y would be moved to the other side} \\ 75-6=4y+4y+3.5y \\ \text{Note that when a positive value moves to the other side of the equation} \\ It\text{ becomes a negative value, and vice versa} \\ 69=11.5y \\ \text{Divide both sides by 11.5} \\ \frac{69}{11.5}=\frac{11.5y}{11.5} \\ 6=y \end{gathered}[/tex]

The answer is y equals 6

Evaluate the expression when x = 32 and y = 2.

x/14 A. 1/16
B.16/21
D.2
C.4

Answers

Answer:

I think its 16/21

Step-by-step explanation:

Answer:

2

Step-by-step explanation:

Given x = 14, y = 2

x/14

Void "y" because it is not in this equation.

= x/14

32/14

= 2.2

≈ 2

A new cell phone costs $108.99 in the store. What would your total cost be if the sale tax is 7.5% ? Round your answer to the nearest cent, if necessary.

Answers

to calculate the tax we need to multiply the % by the price of the cellphone

7.5%=0.075

108.99*0.075=8.17

and the total cost is:

$108.99 + $8.17= 117.16

So the answer is: $

Multiple the binomials (simplify) (y-4)(y-8)

Answers

Given

[tex](y-4)(y-8)[/tex]

Simplify as shown below

[tex]\begin{gathered} (y-4)(y-8)=y(y-8)-4(y-8)=y^2-8y-4y+(-4)(-8)=y^2-12y+32 \\ \Rightarrow(y-4)(y-8)=y^2-12y+32 \end{gathered}[/tex]

The answer is y^2-12y+32

help me please asap!!!

Answers

The slope of the function is 1/2 and the y - intercept is 2

The standard form of slope-intercept form of line is y = mx + b

where , m is slope of line

and b is y-intercept.

Observing the graph ,

we can say Linear function also passes through two points

At (4,0) on x-axis and at (0,2) on y-axis and

also , the graph is making right angles triangle at (0,0)

Slope of the function = m = Tan∅

Tan∅ = Perpendicular of right triangle / base of triangle

Perpendicular of triangle = 2 unit

and base = 4 unit

Tan∅ = 2/4 = 1/2

Therefore , slope of line = 1/2

equation of line : y = 1/2 x + b

This line is passing through (0,2)

2 = 1/2(0) + b

b = 2

Therefore , the y-intercept = 2

Hence , the equation of line = y = 1/2 x + 2

To know more about Slope of the line here ,

https://brainly.com/question/14511992

#SPJ1

Solve the following system of linear equations by graphing:4.Ex+3y54 9- 5858615+cilo=Answer 2 PointsKeypadKeyboard ShortcutsGraph the linear equations by writing the equations in slope-intercept form:y =Ixty =IxtIdentify the appropriate number of solutions. If there is a solution, give thepoint:O One SolutionO No SolutionO Infinite Number of Solutions

Answers

We have a system of equations:

[tex]\begin{gathered} -\frac{4}{5}x+3y=-\frac{58}{5} \\ \frac{4}{3}x+\frac{9}{5}y=\frac{86}{15} \end{gathered}[/tex]

We have to write the equations in slope-intercept form.

We start with the first equation:

[tex]\begin{gathered} -\frac{4}{5}x+3y=-\frac{58}{5} \\ 3y=-\frac{58}{5}+\frac{4}{5}x \\ 5\cdot3y=4x-58 \\ 15y=4x-58 \\ y=\frac{4}{15}x-\frac{58}{15} \end{gathered}[/tex]

For the second equation we get:

[tex]\begin{gathered} \frac{4}{3}x+\frac{9}{5}y=\frac{86}{15} \\ \frac{9}{5}y=\frac{86}{15}-\frac{4}{3}x \\ y=\frac{5}{9}\cdot\frac{86}{15}-\frac{5}{9}\cdot\frac{4}{3}x \\ y=\frac{86}{9\cdot3}-\frac{20}{27}x \\ y=-\frac{20}{27}x+\frac{86}{27} \end{gathered}[/tex]

To graph the equations we need two points. We can easily identify the y-intercept from the equations, but we have to identify one more point for each equation.

We can give a value to x and find the corresponding value of y.

Then, for example we can calculate y for x = 1 in the first equation:

[tex]\begin{gathered} y=\frac{4}{15}(1)-\frac{58}{15} \\ y=\frac{4}{15}-\frac{58}{15} \\ y=-\frac{54}{15} \end{gathered}[/tex]

Then, for the first equation we know the points (0, -58/15) and (1, -54/15).

For the second equation we can do the same, by giving a value of 1 to x (NOTE: we can give any arbitrary value to x, it does not have to be the same for both equations) and calculate y:

[tex]\begin{gathered} y=-\frac{20}{27}(1)+\frac{86}{27} \\ y=-\frac{20}{27}+\frac{86}{27} \\ y=\frac{66}{27} \end{gathered}[/tex]

Now we know the points of the second equation: (0, 86/27) and (1, 66/27).

With such fractions we can not make an accurate graph in paper, as they don't match the divisions of the grid.

We can use approximate decimals values for the fractions and graph the points.

The approximations for the first equation are:

[tex]\begin{gathered} (0,-\frac{58}{15})\approx(0,-3.9) \\ (1,-\frac{54}{15})=(1,-3.6) \end{gathered}[/tex]

and for the second equation:

[tex]\begin{gathered} (0,\frac{86}{27})\approx(0,3.2) \\ (1,\frac{66}{27})\approx(1,2.4) \end{gathered}[/tex]

We can then graph the equations as:

If we graph the equations with the exact points, we get an intersection point at (7,-2).

This intersection is the unique solution to both equations at the same time, so it is the only solution to the system of equations.

Answer:

The equations in slope-intercept form are:

y = 4/15 x + (-58/15)

y = -20/27 * x + 86/27

The system has only one solution: (7, -2).

Is this correct? If not can u show me how to do it

Answers

Given

[tex]k(x+y)=2x-4[/tex]

Notice that it is the equation of a line.

Then, since it crosses (10,-2), set x=10 and y=-2 in the given equation, as shown below

[tex]\begin{gathered} x=10,y=-2 \\ \Rightarrow k(10-2)=2*10-4 \\ \Rightarrow k(8)=20-4 \\ \Rightarrow k=\frac{16}{8}=2 \\ \Rightarrow k=2 \end{gathered}[/tex]

Thus, the answer is k=2.

I’m stuck on how to verify number 7 and how to find the possible value for sin theta

Answers

Given:

There are given the trigonometric function:

[tex]sec^2\theta cos2\theta=1-tan^2\theta[/tex]

Explanation:

To verify the above trigonometric function, we need to solve the left side of the equation.

So,

From the left side of the given equation:

[tex]sec^2\theta cos2\theta[/tex]

Now,

From the formula of cos function:

[tex]cos2\theta=cos^2\theta-sin^2\theta[/tex]

Then,

Use the above formula on the above-left side of the equation:

[tex]sec^2\theta cos2\theta=sec^2\theta(cos^2\theta-sin^2\theta)[/tex]

Now,

From the formula of sec function:

[tex]sec^2\theta=\frac{1}{cos^2\theta}[/tex]

Then,

Apply the above sec function into the above equation:

[tex]\begin{gathered} sec^2\theta cos2\theta=sec^2\theta(cos^2\theta-s\imaginaryI n^2\theta) \\ =\frac{1}{cos^2\theta}(cos^2\theta-s\mathrm{i}n^2\theta) \\ =\frac{(cos^2\theta-s\mathrm{i}n^2\theta)}{cos^2\theta} \end{gathered}[/tex]

Then,

[tex]\frac{(cos^{2}\theta- s\mathrm{\imaginaryI}n^{2}\theta)}{cos^{2}\theta}=\frac{cos^2\theta}{cos^2\theta}-\frac{sin^2\theta}{cos^2\theta}[/tex]

Then,

From the formula for tan function:

[tex]\frac{sin^2\theta}{cos^2\theta}=tan^2\theta[/tex]

Then,

Apply the above formula into the given result:

So,

[tex]\begin{gathered} \frac{(cos^{2}\theta- s\mathrm{\imaginaryI}n^{2}\theta)}{cos^{2}\theta}=\frac{cos^{2}\theta}{cos^{2}\theta}-\frac{s\imaginaryI n^{2}\theta}{cos^{2}\theta} \\ =1-\frac{s\mathrm{i}n^2\theta}{cos^2\theta} \\ =1-tan^2\theta \end{gathered}[/tex]

Final answer:

Hence, the above trigonometric function has been proved.

[tex]sec^2\theta cos2\theta=1-tan^2\theta[/tex]

From the diagram below, we can tell that Δ ABC is similar to ____.

Answers

From the diagram, we get that

[tex]\measuredangle ABD\cong\measuredangle ADB.[/tex]

By the reflexive property of congruence, we know that:

[tex]\measuredangle A\cong\measuredangle A.[/tex]

Therefore, by the Angle-Angle criterion:

[tex]\Delta ABC\sim\Delta ADB.[/tex]Answer: [tex]\Delta ADB.[/tex]

If ∠2 = 50° and lines a and b are parallel, which of the following angles cannot be determined? *-∠1-∠3-∠4-∠8-None of the above

Answers

If you have the angle 2, you can deduce its the supplement,which is the angle 4. In this case the supplement is equal to 130°.

Angle 2 and angle 6 are corresponding angles, so they have the same measure.

Angle 4 and angle 8 are corresponding, so they have the same measure.

Angle 2 and angle 1 are vertically opposite angles, so they have the same measure.

Angle 5 and angle 6 are vertically opposite angles, so they have the same measure.

Angle 7 and angle 8 are vertically opposite angles, so they have the same measure.

Angle 4 and angle 3 are vertically opposite angles, so they have the same measure.

Therefore the answer is None of the above.

using the rule of s-14 - (-2) = -12

Answers

We will have:

[tex]-14-(-2)=-12\Rightarrow-14+2=-12\Rightarrow-12=-12[/tex]

The coordinates of the midpoint of GH are M(-2,5) and the coordinates of one endpoint are H(-3, 7).
The coordinates of the other endpoint are(
).
Echeck
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What are the coordinates of the other endpoint

Answers

EXPLANATION :

From the problem, we have segment GH and the midpoint is M(-2, 5).

One of the endpoints has coordinates of H(-3, 7)

and we need to find the coordinates of G(x, y)

The midpoint formula is :

[tex]M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})[/tex]

where (x1, y1) are the coordinates of G

(x2, y2) = (-3, 7) are the coordinates of H

and (-2, 5) are the coordinates of the midpoint.

Then :

[tex](-2,5)=(\frac{x+(-3)}{2},\frac{y+7}{2})[/tex]

We can equate the x coordinate :

[tex]\begin{gathered} -2=\frac{x+(-3)}{2} \\ \\ \text{ cross multiply :} \\ -2(2)=x-3 \\ -4=x-3 \\ -4+3=x \\ -1=x \\ x=-1 \end{gathered}[/tex]

then the y coordinate :

[tex]\begin{gathered} 5=\frac{y+7}{2} \\ \\ \text{ cross multiply :} \\ 5(2)=y+7 \\ 10=y+7 \\ 10-7=y \\ 3=y \\ y=3 \end{gathered}[/tex]

Now we have the point (-1, 3)

ANSWER :

The coordinates of the other endpoint are G(-1, 3)

Section 5.2-12. Solve the following system of equations by substitution or elimination. Enter your answer as (x,y).3x-3y = -6-x+2y = 8

Answers

the Given the simultaneous equations

[tex]\begin{gathered} 3x-3y=-6\text{ ------(1)} \\ -x+2y=8\text{ -------(2)} \end{gathered}[/tex]

Solving the above equations by substitution method:

Step 1:

From equation 2, make x the subject of the formula

[tex]\begin{gathered} -x+2y=8 \\ \text{making x the subject of formula, we have} \\ x=2y-8\text{ -------(3)} \end{gathered}[/tex]

Step 2:

Substitute equation 3 into equation 1

[tex]\begin{gathered} \text{From equation,} \\ 3x-3y=-6 \\ \text{Thus, we have} \\ 3(2y-8)-3y=-6 \\ \text{opening the brackets, we have} \\ 6y-24-3y=-6 \\ \text{collecting like terms, we have} \\ 6y-3y=-6+24 \\ 3y=18 \\ \text{divide both sides by the coefficient of y.} \\ \text{The coefficient of y is 3. Thus,} \\ y=\frac{18}{3}=6 \end{gathered}[/tex]

Step 3:

Substitute the value of y in either equation 1 or 2.

[tex]\begin{gathered} \text{From equation 2,} \\ -x+2y=8 \\ \text{Thus,} \\ -x+2(6)=8 \\ -x+12=8 \\ \text{collecting like terms, we have} \\ -x=8-12 \\ -x=-4 \\ x=4 \end{gathered}[/tex]

Thus, the values of (x, y) are (4, 6)

Write the percent as decimal 49%

Answers

Solution;

Given: The given number in percentage is 49 %

Required: Decimal value of given percentage.

Explanation:

Convert percentage into decimal as follows:

[tex]49\text{ \%=}\frac{49}{100}[/tex]

[tex]49\text{ \%=0.49}[/tex]

Therefore, the required answer is 0.49

Final answer: The de

How do I find the restrictions on x if there are any? [tex] \frac{1}{x - 1} = \frac{5}{x - 10} [/tex]

Answers

We have the expression:

[tex]\frac{1}{x - 1}=\frac{5}{x - 10}[/tex]

When we have rational functions, where the denominator is a function of x, we have a restriction for the domain for any value of x that makes the denominator equal to 0.

That is because if the denominator is 0, then we have a function f(x) that is a division by zero and is undefined.

If we have a value that makes f(x) to be undefined, then this value of x does not belong to the domain of f(x).

Expression:

[tex]\begin{gathered} \frac{1}{x-1}=\frac{5}{x-10} \\ \frac{x-1}{1}=\frac{x-10}{5} \\ x-1=\frac{x}{5}-\frac{10}{5} \\ x-1=\frac{1}{5}x-2 \\ x-\frac{1}{5}x=-2+1 \\ \frac{4}{5}x=-1 \\ x=-1\cdot\frac{5}{4} \\ x=-\frac{5}{4} \end{gathered}[/tex]

Answer: There is no restriction for x in the expression.

How to find the diagonal side one triangle like the measure with the Pythagorean Theorem

Answers

How to find the diagonal side one triangle like the measure with the Pythagorean Theorem​

see the attached figure to better understand the p

Given f(x)=2x-1 and g(x) =x^2 -2A) f(5)B) f(g(3))C) f(a+1) - f(a)D) g(2f(-1))E) g(x+h) -g(x)/h

Answers

Answer:

2x + h

Explanation:

Given the following functions

f(x) = 2x - 1

g(x) = x^2 - 2

We are to simplify the expressionn:

[tex]\frac{g(x+h)-g(x)}{h}[/tex]

Substitute the given functions into the expression and simplify

[tex]\begin{gathered} \frac{\lbrack(x+h)^2-2\rbrack-(x^2-2)}{h} \\ \frac{\lbrack\cancel{x^2}^{}+2xh+h^2-\cancel{2}-\cancel{x^2}^{}+\cancel{2}}{h} \\ \frac{2xh+h^2}{h} \end{gathered}[/tex]

Factor out "h" from the numerator to have:

[tex]\begin{gathered} \frac{\cancel{h}(2x+h)}{\cancel{h}} \\ 2x+h \end{gathered}[/tex]

Hence the simplified form of the expression is 2x + h

Other Questions
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