What is the symbol for mean in stats?

Answer: x^6+6x^4+8x^2+2

Step-by-step explanation:

Since g comes after f then you will take g's equation and plug it into the f equation so it would turn out to be if you plugged it in

(x^2+2)^3-4(x^2+2)+2 which would equal x^6+6x^4+8x^2+2

P(A∩B) = 0 (the probability of both events occurring is zero).

P(A|B) = 0.

No. The student's argument that mutually exclusive events are independent is incorrect.

If events A and B are mutually exclusive, then by definition they cannot occur at the same time.

Thus, P(A∩B) = 0 (the probability of both events occurring is zero).

Since events A and B are mutually exclusive, they cannot be independent. In fact, if two events are mutually exclusive, it means that the occurrence of one event precludes the occurrence of the other event, so they are completely dependent on each other.

To calculate P(A|B), we can use the formula:

P(A|B) = P(A∩B) / P(B)

Since P(A∩B) = 0 and P(B) = 0.70, we have:

P(A|B) = 0 / 0.70 = 0

This means that if we know event B has occurred (i.e., P(B) > 0), then the probability of event A occurring (i.e., P(A|B)) is zero. Therefore, events A and B are not independent.

The student's argument that mutually exclusive events are independent is incorrect. Mutually exclusive events are by definition dependent events.

In general, the concept of mutually exclusive events and the concept of independent events are not the same. Mutually exclusive events are dependent events, while independent events are events that do not depend on each other.

In this problem, the fact that events A and B are mutually exclusive means that they are dependent events, and this is reflected in the probability calculation for P(A|B), which is zero. Therefore, we can conclude that mutually exclusive events are not independent events.

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Answer:

11/27

Step-by-step explanation:

55/135 (divide by 5)

11/27

:Answer: $90000

i = prt (interest = principle x rate x time)

i = 60000 x 0.025 x 20

i = $30000 interest

b = p + i (balance = principle + interest)

b = 60000 + 30000

b = $90000 balance

Based on the graph of this exponential function, the domain and range are as follows;

Domain = {-∞, ∞}.

Range = {0, ∞}.

In Mathematics, a domain is the set of all real numbers for which a particular function is defined.

Additionally, the vertical extent of any graph of a function represents all range values and they are always read and written from smaller to larger numerical values, and from the bottom of the graph to the top.

By critically observing the graph shown in the image attached above, we can reasonably and logically deduce the following domain and range:

Domain = {-∞, ∞}.

Range = {0, ∞} or {y|y > 0}.

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Using the trapezoid midsegment theorem, the length of the midsegment of trapezoid PQRS is 11 units,

The Trapezoid Midsegment Theorem states that the length of the midsegment of a trapezoid is equal to the average of the lengths of the bases (the parallel sides). Mathematically, this can be expressed as:

length of midsegment = (length of base₁ + length of base₂) / 2

In the case of trapezoid PQRS, the bases are QR and PS.

The length of QR is 8 units and the length of PS is 14 units.

Therefore, the length of the midsegment can be calculated as:

length of midsegment = (8 + 14) / 2

length of midsegment = 22 / 2

= 11 units

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The standard form of the Polynomial would be [tex]x^{4}[/tex]  -x³ /2 -8

The term "polynomial" refers to a mathematical statement of one or more algebraic terms in which the variables are all of non-negative integer powers. Variables, fixed values, and exponents are all present in the terms. The degree 'n' standard form polynomial is Standard form is one method of expressing a polynomial.

We must consider each term's degree in order to write any polynomial in standard form.

Given the polynomial -8 -x³ /2 -xx³ with the highest degree comes first, followed by x raise to power 4, and then the constant, -8, comes last.

Thus, we would have:

[tex]x^{4}[/tex]  -x³ /2 -8

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Answer:

y+75°=180°[co interior angle]

or,y=180-75

•°•y=105

again,

y=z(alternate angle)

•°•z=105°

now

z+x+68°=180(sum of angle of triangle is 180)

or,105+68+x=180

or,x+173=180

or,x=180-173

•°•x=7

The midpoints of the given quadrilateral, AB = CD and AD = BC, are equal.

Let P represent the quadrilateral's vertices (-4,2) Q (2,6): Let A, B, C, and D be the midpoints of P(-4,2) and Q. R(8,5) and S (9,-7) (2,6)

Consequently, A's coordinates are

[tex]$\left(\frac{-4+2}{2}, \frac{2+6}{2}\right)=\left(\frac{-2}{2}, \frac{8}{2}\right)=(-1,4)$[/tex]

Coordinates of B are

[tex]$\left(\frac{2+8}{2}, \frac{6+5}{2}\right)=\left(\frac{10}{2}, \frac{11}{2}\right)=\left(5, \frac{11}{2}\right)[/tex]

Coordinates of c are

[tex]$\left(\frac{8+9}{2}, \frac{5-7}{2}\right)=\left(\frac{17}{2}, \frac{-2}{2}\right)=\left(\frac{17}{2},-1\right)[/tex]

coordinates of D are [tex]$\left(\frac{9-4}{2}, \frac{7+2}{2}\right)=\left(\frac{5}{2}, \frac{-5}{2}\right)$[/tex]

Distance between A and B

[tex]$& d(A, B)=\sqrt{(-1-5)^2+\left(4-\frac{11}{2}\right)^2}=\sqrt{36+\frac{9}{4}}=\sqrt{\frac{153}{4}} \\[/tex]

Simplify,

[tex]$& d(C, D)=\sqrt{\left(\frac{17}{2}-\frac{5}{2}\right)^2+\left(-1+\frac{5}{2}\right)^2}=\sqrt{36+\frac{9}{4}}=\sqrt{\frac{153}{4}} \\[/tex]

Then,

[tex]$& d(A, D)=\sqrt{\left(-1-\frac{5}{2}\right)^2+\left(4+\frac{5}{2}\right)^2}=\sqrt{\frac{49}{4}+\frac{169}{4}}=\sqrt{\frac{218}{4}} \\[/tex]

We get,

[tex]$& d(B, C)=\sqrt{\left(5-\frac{17}{2}\right)^2+\left(\frac{11}{2}+1\right)^2}=\sqrt{\frac{49}{4}+\frac{169}{4}}=\sqrt{\frac{218}{4}}[/tex]

As a result, the given quadrilateral's midpoints, AB = CD and AD = BC, are equal.

It is a parallel gramme as a result.

The complete question is :

A quadrilateral has the vertices at the point (−4,2),(2,6),(8,5) and (9,−7). Show that the mid-point of the sides of this quadrilateral are the vertices of a parallelogram.

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The note Micah made for the test on right triangles is correct by the Pythagoras rule.

The Pythagoras rule states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 

so by Pythagoras rule we can write:

For the first right triangle;

(x/√2)² + (x/√2)² = x²...(1)

from the left hand side of equation (1): (x/√2)² + (x/√2)²

x²/2 + x²/2

2(x²)/2 = x² {the right hand side of equation (1)}

and for the second right triangle;

(x√3/2)² + (x/2)² = x²...(2)

from the left hand side of equation (2): (x√3/2)² + (x/2)²

x²(3)/4 + x²/4

3x²/4 + x²/4

(3x² + x²)/4

4x²/4 = x² {the right hand side of equation (2)}

Therefore by Pythagoras rule the note Micah made for the test on right triangles is correct

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The ordered pair that show a combination of 12 coins from the graph is (10, 2)

from the question, we have the following parameters that can be used in our computation:

The graph

The purple line represents the combination of 12 coine

Using the above as a guide, we have the following:

The equation is

x + y = 12

From the list of options, we have

(10, 2): 10 + 2 = 12

Hence, the ordered pair is (10, 2)

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Answer: The type of function that best models the data would depend on the actual change in the number of students and the specific data points for each year. However, if the change in the number of students was the same for each time period, a linear function would be a good candidate for modeling the data. This function would have the form y = mx + b, where m represents the rate of change (the change in the number of students over a certain time period) and b is the y-intercept (the number of students at the starting year).

Step-by-step explanation:

The point (-0.5, 0.866) represents the angle -4pi/3 graphed on the unit circle.

For an angle [tex]\theta[/tex] the unit circle is a circle with radius 1 containing the following set of points: [tex](\cos{\theta}, \sin{\theta})[/tex].

The angle for this problem is given as follows:

-4pi/3.

Hence the sine and the cosine are given as follows:

Hence the point (-0.5, 0.866) represents the angle -4pi/3 graphed on the unit circle.

(the equation of the unit circle is x² + y² = 1).

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Answer:

Like the other solver, I believe they are correct because I arrived at the same solution:

3/4w <= 225

Multiplying both sides by 4/3 to isolate w:

w >= (225 * 4)/3

w >= 300

Solution is w >= 300. In order to meet the inequality, the value of w must be equal to or larger than 300.

Answer:

Step-by-step explanation:

A distribution is said to be skewed to the right (also known as positively skewed) if it has a longer tail on the positive side of the mean or median, indicating that the majority of the data values are on the lower end and there are a few large values on the higher end. This results in a mean that is larger than the median.

In other words, the right skew occurs when the tail of the distribution is located to the right side of the peak, which pulls the mean to the right as well. The presence of outliers or extreme values on the right side of the distribution can cause right skew.

Right-skewed distributions are common in data sets that represent measurements or quantities that have a minimum value of zero or a natural lower limit, but can increase without bound, such as income, wealth, and height.

In summary, when a distribution is skewed to the right, it has a higher mean and lower median compared to a symmetrical distribution, and the majority of the data is concentrated on the left side.

Answer:

See below

Step-by-step explanation:

[tex]I_1=[1],\, I_2=\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right],\,I_3=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right],\,...[/tex] and so on are examples of identity matrices.

The multiplicative inverse of a 2x2 matrix is [tex]\displaystyle A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right][/tex] where ad-bc represents the determinant. Multiplying this with the original matrix will produce the identity matrix as described previously. Usually, you'll mostly deal with 2x2 matrices when it comes to this topic, but don't worry too much.

The statement "a quadrilateral is a square if and only if it is a rectangle" made by Ruby is a true biconditional statement in mathematical terms.

A quadrilateral is a four-sided shape that can have different properties based on its sides and angles.

Ruby says that a quadrilateral is a square if and only if it is a rectangle. This statement can be written as:

(Q is a square) if and only if (Q is a rectangle)

To understand if this statement is a true biconditional statement, we need to examine the definition of a square and a rectangle and how they relate to each other.

A square is a type of quadrilateral where all four sides are equal in length and all four angles are equal (90 degrees). In other words, a square is a rectangle that has four equal sides.

A rectangle is a type of quadrilateral where all four angles are equal to 90 degrees. This means that the opposite sides are parallel and equal in length.

Based on the definitions, we can conclude that the statement "a quadrilateral is a square if and only if it is a rectangle" is a true biconditional statement.

This means that if a quadrilateral is a square, it is also a rectangle and vice versa. If a quadrilateral is not a square, it cannot be a rectangle.

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Answer:

Below

Step-by-step explanation:

3 x^2 = 48      divide both sides of the equation by 3 to get

x^2 = 16         Now, sqrt both sides

x = + 4  or  -4

The length of the diameter of the circle is 8 cm.

The diameter of a circle is the distance measured through its centre. The radius of a circle is the distance from the center to any point on the edge. The diameter is divided by the radius.

Given:

Radius = 4 cm

As, the radius is the point which starts from center  and reach to point on the surface of the Circle.

and, the Diamter is the two times of the radius.

So, Diameter = 2 x radius

D= 2 x 4

diameter = 8 cm

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By evaluating the linear equation we can see that the complete table is:

x|  -4   -2  0  2  4

y| 17    13  9   5  1

Here we have a linear function:

y = -2x + 9

And we want to complete the given table, to do so, we just need to evaluate the linear function in the given values of x.

if x = -2

y = -2*-2 + 9 = 4 + 9 = 13

if x = 0

y = -2*0 + 9 = 9

if x = 2

y = -2*2 + 9 = 5

if x = 4

y = -2*4 +9 = 1

Then the complete table is:

x|  -4   -2  0  2  4

y| 17    13  9   5  1

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Answer:

No, Tony should have rounded up to 38.

Step-by-step explanation:

Tony solved the inequality correctly; however, he rounded incorrectly.

8c+200≥500

Subtract 200 from both sides

8c≥300

Divide each side by 8

c≥37.5

Tony cannot wash 37.5 cars, because in the context it is impossible to wash half of a car. We round the answer up because rounding down would give us a value less than $500 (the inequality is greater than or equal to).

Therefore, the answer is 38. Tony is incorrect.

The number of freshmen that are students is 738

Let y represent the number of students that are freshmen

The freshmen class is made up of 480 female students

This number represents 65% of the freshmen class

Therefore the total number of students that are freshmen can be calculated as follows

y × 65/100= 480

65y/100= 480

Cross multiply both sides

65y= 480 × 100

65y= 48000

y= 48000/65

y= 738.4

≅ 738

Hence 738 students are freshmen

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The  perimeter of the painting 7 feet.

Perimeter is the distance around the edge of a shape. To find the perimeter by adding up the side lengths of various shapes.

Given:

A rectangle with side lengths y and z has perimeter 2y+2z.

Elizabeth bought a rectangular painting whose length = 2 ft and width = 1.5 ft.

So, the  perimeter of the painting

= 2l+ 2w

= 2(2)+ 2(1.5)

= 4 + 3

= 7 feet

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The equation for A is A=4b(t+D) from t = A/4b-D.

Two or more expressions with an Equal sign is called as Equation.

The given equation is t = A/4b-D

t equal to A by four times of b minus D

The variables in the equation are A, b and D.

The formula is given for the variable t.

We need to change the formula to find the value of A.

We need to solve for A.

Add D on both sides

t+D=A/4b-D+D

The positive and negative D on right side of the equation will be cancelled and left with A/4b.

t+D/1=A/4b

Now apply cross multiplication

A=4b(t+D)

Hence, the equation for A is A=4b(t+D) from the t = A/4b-D.

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The equation of the circle graphed is given as follows:

(x + 3)² + (y - 4)² = 41.

The equation of a circle of center [tex](x_0, y_0)[/tex] and radius r is given by:

[tex](x - x_0)^2 + (y - y_0)^2 = r^2[/tex]

The center in this problem has the coordinates given as follows:

(-3,4).

Hence:

(x + 3)² + (y - 4)² = r².

x = -1 and y = 9 is a point on the circumference of the circle, hence the radius of the circle is obtained as follows:

r² = (-1 + 3)² + (9 - 4)²

r² = 41.

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The requried measure of length JK on the coordinate plane is 10 units.

The distance between two points on the coordinate plane can be evaluated by the distance formula as,
D = √((x₂ - x₁)^2 + (y₂ - y₁)²)

Here,
The coordinates of the two points J(-7, -7) and K(-7, 3) suggest that they lie on a vertical line, where the x-coordinate is constant (-7) and only the y-coordinate changes. The distance between these points is equal to the difference in their y-coordinates.
Therefore, the length of JK is,
JK = |y-coordinate of K - y-coordinate of J|
    = |3 - (-7)| = |3 + 7| = 10

So, the length of JK is 10 units.

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Answer: Robert has 35 cans of paint - 30 cans of paint that he has already used = 35 - 30 = 5 cans of paint left.

Since each section of trim takes 1/2 can of paint, Robert can paint 2 sections of trim with 1 can of paint.

Therefore, Robert can paint 5 cans of paint * 2 sections of trim per can = 10 sections of trim.

The inequality can be written as 10 <= sections of trim <= 10.

In words, the number of sections of trim that Robert can paint is less than or equal to 10.

On a number line, this solution can be represented as follows:

[9, 10, 11, ...]

 ^

 |

sections of trim

The solution makes sense because Robert has only 5 cans of paint left, so it is not possible for him to paint more than 10 sections of trim.

Step-by-step explanation:

[tex]\huge\text{Hey there!}[/tex]

[tex]\mathsf{\dfrac{3}{4} + \dfrac{1}{2}}[/tex]

[tex]\mathsf{= \dfrac{3}{4} + \dfrac{1\times2}{2\times2}}[/tex]

[tex]\mathsf{= \dfrac{3}{4} + \dfrac{2}{4}}[/tex]

[tex]\mathsf{= \dfrac{3 + 2}{4 + 0}}[/tex]

[tex]\mathsf{= \dfrac{5}{4}}[/tex]

[tex]\mathsf{= 1 \dfrac{1}{4}}[/tex]

[tex]\huge\text{Therefore your answer should be:}[/tex]

[tex]\huge\boxed{\mathsf{\dfrac{5}{4}\ or \ 1 \dfrac{1}{4}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]