In an experiment, the probability that event A occurs is 6/7, the probability that event B occurs is 4/5, and the probability that event A and B both occur is 2/3. what is the probability that A occurs given that B occurs?

Answer:

When solving a compound inequality, you can express the solution in two different ways: using a union or intersection.

Using a union:

A union represents the solutions that belong to either one inequality or the other, or both. You can express a union using the symbol "or," or by writing the solutions for each inequality separately and combining them using a "union" symbol (∪). For example, if you have the compound inequality "x > 2 or x < -3," you can express the solution as "x ∈ (-3, 2) ∪ (-∞, -3) ∪ (2, ∞)."

Using an intersection:

An intersection represents the solutions that belong to both inequalities. You can express an intersection using the symbol "and," or by writing the solutions for each inequality separately and combining them using an "intersection" symbol (∩). For example, if you have the compound inequality "x > 2 and x < -3," you can express the solution as "x ∈ (-3, 2) ∩ (-∞, -3) ∩ (2, ∞)."

Which method you use to express the solution will depend on the specific inequalities and the type of solutions you are looking for. It is important to carefully consider the inequality symbols (>, <, ≥, ≤) and the use of "and" or "or" in order to determine the correct solution.

When solving a compound inequality, the solution can be expressed in one of two ways: as an interval or as a union of intervals.

Inequality is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are not equal.

An interval is a set of all real numbers between two given numbers, and is written in the form (a, b) or [a, b], where a and b are real numbers, and a < b. If a and b are included in the interval, then the interval is written as [a, b]; if a and b are not included in the interval, then the interval is written as (a, b).

A union of intervals is a combination of two or more intervals, and is written in the form (a, b) ∪ (c, d) or [a, b] ∪ [c, d], where a, b, c, and d are real numbers, and a < b and c < d.

When solving a compound inequality, it is important to first simplify the inequality by combining like terms, isolating the variable on one side of the inequality, and combining any separate inequalities into a single inequality. Then, the solution can be found by graphing the inequality on a number line, and shading the portion of the number line that represents the solution.

Once the solution is found the solution can be expressed either as an interval or as a union of intervals depending on the inequality. For example, if the inequality is x>1 and x<3, then the solution would be expressed as (1,3) and this is an example of an interval. If it's x>1 and x≥3 then the solution would be expressed as (1,3] U [3,∞) and this is an example of union of intervals.

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Answer:

x=17 degrees

Step-by-step explanation:

We know that a right angle is equal to 90 degrees. Therefore, two angles between a right angle have to equal 90. Therefore, 73+x=90. Isolate x and get x=90-73. x=17. Therefore, x=17 degrees.

If this helps please mark as brainliest

Answer:

the answer is

B. 145° 12' 0"

Answer:

8, 45, 18, 8.75 pretty sure

Step-by-step explanation:

cross multiplying: when a/b=x/c you type in a*c/b to get x in calculator

Answer:

$432,189

Step-by-step explanation:

$1,000,000 - $567,811 = $432,189

The design of a simulation that would allow Ian to determine how likely it would

be to guess all answers correctly on this quiz can be Illustrated through a coin.

Ian can design a simulation to determine the likelihood of guessing all answers correctly on the quiz by using a fair coin. He can flip the coin four times, with each flip representing a guess on the quiz. If the coin lands heads up, he marks it as a true answer, and if it lands tails up, he marks it as a false answer.

He can repeat this process a large number of times (e.g. 1000) and count the number of times that the sequence T, T, F, T is generated. The proportion of times that this sequence is generated out of the total number of simulations will be an estimate of the probability of guessing all answers correctly on the quiz by chance.

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Answer:

research it many my answer will be wrong

3

Step-by-step explanation:

-3 square is 9, adding 3 you will have 12. 10-6 is 4. so dividing 12 by 4 would give you 3.

Answer:

1/8 pound per person

Step-by-step explanation:

Answer:

Step-by-step explanation:

distance = [tex]\sqrt{(-5/2+1/2)^{2} +(9/4+3/4)^{2} }=\sqrt{13}[/tex]=3.61

Answer: A. Quadrilateral

Step-by-step explanation:

The name for a polygon with 4 sides is a quadrilateral. Quadrilateral is a two-dimensional shape with four straight sides and four angles. The most common type of quadrilateral is the rectangle, which has four right angles. Other types of quadrilaterals include squares, parallelograms, trapezoids, and rhombuses.

Answer:

quadrilateral!!

Step-by-step explanation:

look at the beginning of the word, quad. Quad means 4 and is referring to a 4 sided figure.

Answer:

8 = x²

Step-by-step explanation:

Logarithmic form: logₓ(8) = 2

Exponential form: 8 = x²

Answer:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c[/tex]

Step-by-step explanation:

Given

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]

Required

Integrate

We have:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]

Let

[tex]u = x^5[/tex]

Differentiate

[tex]\frac{du}{dx} = 5x^4[/tex]

Make dx the subject

[tex]dx = \frac{du}{5x^4}[/tex]

So, we have:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx[/tex]

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, \frac{du}{5x^4}[/tex]

[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{10}}} \, du[/tex]

Express x^(10) as x^(5*2)

[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{5*2}}} \, du[/tex]

Rewrite as:

[tex]\frac{1}{5} \int\ {\frac{1}{1 + x^{5)^2}}} \, du[/tex]

Recall that: [tex]u = x^5[/tex]

[tex]\frac{1}{5} \int\ {\frac{1}{1 + u^2}}} \, du[/tex]

Integrate

[tex]\frac{1}{5} * \arctan(u) + c[/tex]

Substitute: [tex]u = x^5[/tex]

[tex]\frac{1}{5} * \arctan(x^5) + c[/tex]

Hence:

[tex]\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c[/tex]

Answer:

1/24

Step-by-step explanation:

Solve for y =4

48x=2

Answer:

[tex]A.\ \tan(x) \to 2.\ \sin(x) \sec(x)[/tex]

[tex]B.\ \cos(x) \to 5. \sec(x) - \sec(x)\sin^2(x)[/tex]

[tex]C.\ \sec(x)csc(x) \to 3. \tan(x) + \cot(x)[/tex]

[tex]D. \frac{1 - (cos(x))^2}{cos(x)} \to 1. \sin(x) \tan(x)[/tex]

[tex]E.\ 2\sec(x) \to\ 4.\ \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex]

Step-by-step explanation:

Given

[tex]A.\ \tan(x)[/tex]

[tex]B.\ \cos(x)[/tex]

[tex]C.\ \sec(x)csc(x)[/tex]

[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]

[tex]E.\ 2\sec(x)[/tex]

Required

Match the above with the appropriate identity from

[tex]1.\ \sin(x) \tan(x)[/tex]

[tex]2.\ \sin(x) \sec(x)[/tex]

[tex]3.\ \tan(x) + \cot(x)[/tex]

[tex]4.\ \frac{cos(x)}{1 - sin(x)} + \frac{1 - \sin(x)}{cos(x)}[/tex]

[tex]5.\ \sec(x) - \sec(x)(\sin(x))^2[/tex]

Solving (A):

[tex]A.\ \tan(x)[/tex]

In trigonometry,

[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]

So, we have:

[tex]\tan(x) = \frac{\sin(x)}{\cos(x)}[/tex]

Split

[tex]\tan(x) = \sin(x) * \frac{1}{\cos(x)}[/tex]

In trigonometry

[tex]\frac{1}{\cos(x)} =sec(x)[/tex]

So, we have:

[tex]\tan(x) = \sin(x) * \sec(x)[/tex]

[tex]\tan(x) = \sin(x) \sec(x)[/tex] --- proved

Solving (b):

[tex]B.\ \cos(x)[/tex]

Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1

So, we have:

[tex]\cos(x) = \cos(x) * \frac{\cos(x)}{\cos(x)}[/tex]

[tex]\cos(x) = \frac{\cos^2(x)}{\cos(x)}[/tex]

In trigonometry:

[tex]\cos^2(x) = 1 - \sin^2(x)[/tex]

So, we have:

[tex]\cos(x) = \frac{1 - \sin^2(x)}{\cos(x)}[/tex]

Split

[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{\sin^2(x)}{\cos(x)}[/tex]

Rewrite as:

[tex]\cos(x) = \frac{1}{\cos(x)} - \frac{1}{\cos(x)}*\sin^2(x)[/tex]

Express [tex]\frac{1}{\cos(x)}\ as\ \sec(x)[/tex]

[tex]\cos(x) = \sec(x) - \sec(x) * \sin^2(x)[/tex]

[tex]\cos(x) = \sec(x) - \sec(x)\sin^2(x)[/tex] --- proved

Solving (C):

[tex]C.\ \sec(x)csc(x)[/tex]

In trigonometry

[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]

and

[tex]\csc(x)= \frac{1}{\sin(x)}[/tex]

So, we have:

[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)}[/tex]

Multiply by [tex]\frac{\cos(x)}{\cos(x)}[/tex] --- an equivalent of 1

[tex]\sec(x)csc(x) = \frac{1}{\cos(x)}*\frac{1}{\sin(x)} * \frac{\cos(x)}{\cos(x)}[/tex]

[tex]\sec(x)csc(x) = \frac{1}{\cos^2(x)}*\frac{\cos(x)}{\sin(x)}[/tex]

Express [tex]\frac{1}{\cos^2(x)}\ as\ \sec^2(x)[/tex] and [tex]\frac{\cos(x)}{\sin(x)}\ as\ \frac{1}{\tan(x)}[/tex]

[tex]\sec(x)csc(x) = \sec^2(x)*\frac{1}{\tan(x)}[/tex]

[tex]\sec(x)csc(x) = \frac{\sec^2(x)}{\tan(x)}[/tex]

In trigonometry:

[tex]tan^2(x) + 1 =\sec^2(x)[/tex]

So, we have:

[tex]\sec(x)csc(x) = \frac{\tan^2(x) + 1}{\tan(x)}[/tex]

Split

[tex]\sec(x)csc(x) = \frac{\tan^2(x)}{\tan(x)} + \frac{1}{\tan(x)}[/tex]

Simplify

[tex]\sec(x)csc(x) = \tan(x) + \cot(x)[/tex]  proved

Solving (D)

[tex]D.\ \frac{1 - (cos(x))^2}{cos(x)}[/tex]

Open bracket

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \frac{1 - cos^2(x)}{cos(x)}[/tex]

[tex]1 - \cos^2(x) = \sin^2(x)[/tex]

So, we have:

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \frac{sin^2(x)}{cos(x)}[/tex]

Split

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \frac{sin(x)}{cos(x)}[/tex]

[tex]\frac{sin(x)}{\cos(x)} = \tan(x)[/tex]

So, we have:

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) * \tan(x)[/tex]

[tex]\frac{1 - (cos(x))^2}{cos(x)} = \sin(x) \tan(x)[/tex] --- proved

Solving (E):

[tex]E.\ 2\sec(x)[/tex]

In trigonometry

[tex]\sec(x)= \frac{1}{\cos(x)}[/tex]

So, we have:

[tex]2\sec(x) = 2 * \frac{1}{\cos(x)}[/tex]

[tex]2\sec(x) = \frac{2}{\cos(x)}[/tex]

Multiply by [tex]\frac{1 - \sin(x)}{1 - \sin(x)}[/tex] --- an equivalent of 1

[tex]2\sec(x) = \frac{2}{\cos(x)} * \frac{1 - \sin(x)}{1 - \sin(x)}[/tex]

[tex]2\sec(x) = \frac{2(1 - \sin(x))}{(1 - \sin(x))\cos(x)}[/tex]

Open bracket

[tex]2\sec(x) = \frac{2 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]

Express 2 as 1 + 1

[tex]2\sec(x) = \frac{1+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]

Express 1 as [tex]\sin^2(x) + \cos^2(x)[/tex]

[tex]2\sec(x) = \frac{\sin^2(x) + \cos^2(x)+1 - 2\sin(x)}{(1 - \sin(x))\cos(x)}[/tex]

Rewrite as:

[tex]2\sec(x) = \frac{\cos^2(x)+1 - 2\sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]

Expand

[tex]2\sec(x) = \frac{\cos^2(x)+1 - \sin(x)- \sin(x)+\sin^2(x)}{(1 - \sin(x))\cos(x)}[/tex]

Factorize

[tex]2\sec(x) = \frac{\cos^2(x)+1(1 - \sin(x))- \sin(x)(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]

Factor out 1 - sin(x)

[tex]2\sec(x) = \frac{\cos^2(x)+(1- \sin(x))(1-\sin(x))}{(1 - \sin(x))\cos(x)}[/tex]

Express as squares

[tex]2\sec(x) = \frac{\cos^2(x)+(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]

Split

[tex]2\sec(x) = \frac{\cos^2(x)}{(1 - \sin(x))\cos(x)} +\frac{(1-\sin(x))^2}{(1 - \sin(x))\cos(x)}[/tex]

Cancel out like factors

[tex]2\sec(x) = \frac{\cos(x)}{1 - \sin(x)} +\frac{1-\sin(x)}{\cos(x)}[/tex] --- proved

now, we're assuming this is some rectangular area, and thus is simply the product of both quantities, let's change all mixed fractions to improper fractions firstly.

[tex]\stackrel{mixed}{12\frac{1}{2}}\implies \cfrac{12\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{25}{2}} ~\hfill \stackrel{mixed}{11\frac{3}{4}} \implies \cfrac{11\cdot 4+3}{4} \implies \stackrel{improper}{\cfrac{47}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{25}{2}\cdot \cfrac{47}{4}\implies \cfrac{1175}{8}\implies 146\frac{7}{8}~ft^2[/tex]

Answer:

$36

Step-by-step explanation:

we need to find the cost of 12 items, and the function of determining the cost is C(x)=2x+12, where x is the number of items (the input value of the function)

we know that there are 12 items, so in this function, x=12

substitute x as 12 in the function; the x in C(x) also gets substituted as 12, as 12 is the input value and the value of C(x) is the output

C(12)=2(12)+12

multiply

C(12)=24+12

add

C(12)=36

that means, for 12 items, the cost will be $36

Answer:

(1/6) of the students did not pass.

Ans: (1/6)30 = 5 of the students did not pass

To solve this question, we need to first figure out, in fractional form, the number of students who did not pass. The problem tells us that 5/6 of the students passed. Therefore, we know that 1/6 of the students did not pass, because 5/6 + 1/6 = 1. So, all we need to do is multiply 1/6 by the total number of students:

1/6 * 36 = 6

So, 6 students did not pass.

Answer:

Assuming data is from population, 0.1

Step-by-step explanation:

Answer:

The length of the diagonal is 12.53

Step-by-step explanation:

Since we have a rectangle, the diagonal is the hypotenuse of a right triangle. So we can use Pythagorean Theorem.

a^2 + b^2 = c^2

We know the two legs are 6 and 11, so we can find the hypotenuse (which is the diagonal) see image.

See image.

Answer:

To solve this problem, we can set up the following equation:

Huilan's age + Thomas's age = 79

Let H be Huilan's age and T be Thomas's age. We can then rewrite the equation as:

H + T = 79

We are told that Huilan is 7 years older than Thomas, so we can write the following equation:

H = T + 7

Substituting the second equation into the first equation, we get:

(T + 7) + T = 79

Combining like terms, we get:

2T + 7 = 79

Subtracting 7 from both sides, we get:

2T = 72

Dividing both sides by 2, we get:

T = 36

Therefore, Thomas's age is 36 years.

Step-by-step explanation:

Answer:

let Thomas age be a

there fore, hullans age is a+7

(a+7)+a=79

2a=79-7

2a=72

a=36

Answer:

it will take Johnny 7 hours

Step-by-step explanation:

It was easy, i divided 105 by 15 and got 7

Have a wonderful day :)

Answer:

7 hours

Step-by-step explanation:

15 * x = 105

---          ---

15           7

x=7

Answer:

Hello There!!

Step-by-step explanation:

The answer is 15 as 3a=3×a so 3×5=15.

hope this helps,have a great day!!

~Pinky~

Answer:

it would be 10392

Step-by-step explanation:

10392÷8=1299

Answer:

10392

Step-by-step explanation:

10391 ÷ 8 = 1298.875

Not divisible by 8

10392 ÷ 8 = 1299

Divisible by 8

The given polynomial is a factor of P(x) because P(-5)=0.

The Remainder Theorem states that if a synthetic division of a polynomial by x = a results in a zero remainder, then x = an is a zero of the polynomial (thanks to the Remainder Theorem), and x an is also a factor of the polynomial (courtesy of the Factor Theorem).

The Factor Theorem states that x - c is only a factor of P(x) when P(c) = 0.

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Answer:

1st question :55%

2nd ":122.22%

3rd ":81.82%

Answer:

10 females, 15 males

Step-by-step explanation:

Let the number of females in the class be x. As the number of males is 5 more than the number of females, which is x, the number of males would be shown as x+5.

The number of females plus the number of males equal 25, so we get this equation:

x + (x+5) = 25

x + x + 5 = 25

2x + 5 = 25

2x = 20

x = 10.

So, there are 10 females, and since there are 5 more males than females, there are 10 + 5 = 15 males.

Answer:

8 × 10 + 5 × 1. is the correct answer .STAY BRAINLY

The claim is not correct as both the triangles do not meet the SAS triangle congruence criterion.

If any two sides and angle included between the sides of one triangle are equivalent to the corresponding two sides and the angle between the sides of the second triangle, then the two triangles are said to be congruent by SAS rule.

As from the definition above we can understand that for the triangles to be congruent the sides and the angle included between them should be same.

In the given triangles

The two sides of both the triangle has same length but the angle 30 degree is not the angle included for both the traingles

Therfore we can say that the triangles are not congruent with SAS congruency.

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