The pairs of polygons below are similar. Give the scale factor of figure A to figure B.
1/8
2
1/4
8
1/2
4
Answers
From figure A to figure B, the scale factors are 1/8 and 1/2. To scale shapes in various dimensions, utilize the scale factor.
what is polygon ?A polygon is a closed object in a two-dimensional plane comprised of line segments rather than curves. The word "polygon" is a combination of the words "poly" (which meaning many) and "gon" (means sides). A closed figure must include a minimum of three line segments that connect end to end. A polygon is any three or more-sided closed two-dimensional shape. In addition to more complex designs like a dodecagon with twelve sides, polygons include triangles and squares. Polygons include shapes like quadrilaterals, triangles, hexagons, and pentagons. The number of sides the form has is indicated by the name.
given
the scale factor of figure A to figure B
= smaller dimension / larger dimension
= 2/16
= 1/8
and 4/8
= 1/2
From figure A to figure B, the scale factors are 1/8 and 1/2. To scale shapes in various dimensions, utilize the scale factor.
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Related Questions
someone help me plsss
Answers
x2 - 3x - 5 = 0 has a unique solution of type -11, which is what the equation entails.
what is discriminant ?The "discriminant" is the value that aids in identifying the many types of solutions when using the quadratic formula to solve a problem. It is a unique function of a co-efficient in a mathematical equation (a quadratic equation), and the value it yields provides details about the roots of the equation. Calculating the discriminant involves square rooting the "b" term and subtracting four times the "a" term from the "c" term. Alongside "D," there is a (delta) that represents discrimination. A quadratic equation with two distinct real number solutions is said to have a positive discriminant. A quadratic equation with a discriminant of zero has repeated real number solutions. Both of the solutions are not real numbers, as indicated by a negative discriminant.
given
x2 - 3x - 5 = 0
D=b2−4ac
D= −32−4×1×5
D=92−20
D= −11
x2 - 3x - 5 = 0 has a unique solution of type -11, which is what the equation entails.
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a male animal can weigh up to 8,000 pounds. How many tons is 8,000 pounds
Answers
Answer:
4
Step-by-step explanation:
4x + 2y = 8 and 16x – y = 14 what does x and y equal?
Answers
Answer:
here is ur answer
Step-by-step explanation:
4x +2y = 4
4x = 4 -2y
x = 4/4 -2y
x= -2 y
Find the equation of the normal to the circle, whose equation is given below, at the point (1,2).( x − 3 ) 2 + ( y + 2 ) 2 = 20
Answers
Answer:
The equation of the normal is y = 4 - 2·x
Step-by-step explanation:
The given equation of the circle, is presented as follows;
(x - 3)² + (y + 2)² = 20
The point of the normal of the circle = (1, 2)
The equation of the normal to a circle, x² + y² + D·x + E·y + F = 0 at a point P(x₁, y₁) is given as follows;
[tex]\dfrac{y - y_1}{x - x_1} = \dfrac{2 \cdot y_1 + D}{2 \cdot x_1 + E}[/tex]
Expanding the given equation of the circle, gives;
(x - 3)² + (y + 2)² = x² + y² - 6·x + 4·y + 13 = 20
∴ x² + y² - 6·x + 4·y + 13 - 20 = x² + y² - 6·x + 4·y - 7 = 0
x² + y² - 6·x + 4·y - 7 = 0
∴ x₁ = 1, y₁ = 2, D = -6, E = 4, and F = 13
Which gives;
[tex]\dfrac{y - 2}{x - 1} = \dfrac{2 \times 2 + 4}{2 \times 1 + (-6)} = \dfrac{8}{-4} = -2[/tex]
∴ y - 2 = -2 × (x - 1) = 2 - 2·x
y = 2 - 2·x + 2 = 4 - 2·x
The equation of the normal to the circle with equation (x - 3)² + (y + 2)² = 20, at the point (1, 2) is y = 4 - 2·x
Find a basis for the eigenspace corresponding to each listed eigenvalue of A below.
A = 4 0 -1 14 5 -10 2 0 1 λ=5,2,3
A basis for the eigenspace corresponding to λ = 5 is { }. (Use a comma to separate answers as needed.) A basis for the eigenspace corresponding to λ = 2 is { }. (Use a comma to separate answers as needed.) A basis for the eigenspace corresponding to λ = 3 is . { }. (Use a comma to separate answers as needed.)
Answers
The basis for the eigenspace corresponding to lambda=5,1,4 are None,[tex]\left[\begin{array}{c}-1 \\\frac{1}{2} \\0\end{array}\right][/tex] and [tex]$\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]$[/tex]
[tex]$$A=\left[\begin{array}{ccc}5 & -12 & 10 \\0 & 7 & -3 \\0 & 6 & -2\end{array}\right]$$[/tex]
Eigenspace corresponding to lambda=5,1,4
The eigenspace E_lambda corresponding to the eigenvalue lambda is the null space of the matrix a [tex]\mathrm{A}-(\lambda) \mathrm{I}"[/tex]
for lambda=5
[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-5 \mathrm{I})$$[/tex]
Reducing the matrix A-5I by elementary row operations
[tex]$$\begin{aligned}A-5 I & =\left[\begin{array}{ccc}5-5 & -12 & 10 \\0 & 7-5 & -3 \\0 & 6 & -2-5\end{array}\right] \\& =\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 2 & -3 \\0 & 6 & -7\end{array}\right] \\& \sim\left[\begin{array}{ccc}0 & -12 & 10 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_2 \rightarrow \frac{R_2}{2} \\& \sim\left[\begin{array}{ccc}1 & 0 & -8 \\0 & 1 & -\frac{3}{2} \\0 & 6 & -7\end{array}\right] R_1 \rightarrow R_1+2 R_2\end{aligned}$$[/tex]
[tex]\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 2\end{array}\right] R_3 \rightarrow R_3-6 R_2$$\\\sim\left[\begin{array}{ccc}1 & 0 & -8 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] R_3 \rightarrow \frac{\mathrm{R}_3}{2}$$\\\sim\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & -\frac{3}{2} \\ 0 & 0 & 1\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+8 \mathrm{R}_3$[/tex]
[tex]$\sim\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] R_2 \rightarrow R_2+\frac{2 R_3}{2}$[/tex]
The solutions x of A-5I=0 satisfy x_1=x_2=x_3=0 that is, the null space solves the matrix
[tex]$$\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]
Hence The null space is [tex]\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right] E_5[/tex] has no basis
[tex]$$\begin{aligned}& \text { case: } 2 \\& \text { for } \lambda=1 \\& \mathrm{E}_5=\mathrm{N}(\mathrm{A}-(1) \mathrm{I})\end{aligned}$$[/tex]
we reduce the matrix A-I by elementary row operations as follows.
[tex]$$\begin{aligned}A-1 & =\left[\begin{array}{ccc}5-1 & -12 & 10 \\0 & 7-1 & -3 \\0 & 6 & -2-1\end{array}\right] \\& =\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 6 & -3 \\0 & 6 & -3\end{array}\right] R_1 \rightarrow \frac{R_1}{4} \\& \sim\left[\begin{array}{ccc}1 & -3 & \frac{5}{2} \\0 & 1 & -\frac{1}{2} \\0 & 6 & -3\end{array}\right] R_2 \rightarrow \frac{R_2}{6}\end{aligned}[/tex]
[tex]$$$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 6 & -3\end{array}\right] R_1 \rightarrow R_1+3 R_2$\\$\sim\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right] R_3 \rightarrow R_3-6 R_2$[/tex]
Thus, the solutions x of (A-I) X=0 satisfy
[tex]$\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\ x_2 \\ x_3\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$[/tex]
x_3=t
[tex]$\Rightarrow \mathrm{x}_1=-\mathrm{t}, \mathrm{x}_2=\frac{\mathrm{t}}{2}$[/tex]
[tex]$\vec{x}=\left[\begin{array}{c}-t \\ \frac{t}{2} \\ t\end{array}\right]=\left[\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right] t$[/tex]
The Basis for the nullspace A-I will be: [tex]$\left.\left(\begin{array}{c}-1 \\ \frac{1}{2} \\ 1\end{array}\right]\right)$[/tex]
case:3
lambda=4
[tex]$$\mathrm{E}_5=\mathrm{N}(\mathrm{A}-(4) \mathrm{I})$$[/tex]
we reduce the matrix A-4I by elementary row operations as follows.
[tex]$\begin{aligned} A-4 \mid & =\left[\begin{array}{ccc}5-4 & -12 & 10 \\ 0 & 7-4 & -3 \\ 0 & 6 & -2-4\end{array}\right] \\ & =\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 3 & -3 \\ 0 & 6 & -6\end{array}\right] \\ & \sim\left[\begin{array}{ccc}1 & -12 & 10 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] R_2 \rightarrow \frac{R_2}{3}\end{aligned}$[/tex]
[tex]$\begin{aligned} & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 6 & -6\end{array}\right] \mathrm{R}_1 \rightarrow \mathrm{R}_1+12 \mathrm{R}_2 \\ & \sim\left[\begin{array}{ccc}1 & 0 & -2 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right] \mathrm{R}_3 \rightarrow \mathrm{R}_3-6 \mathrm{R}_2\end{aligned}$[/tex]
Thus, the solutions x of (A-4IX)=0 satisfy
[tex]$$\left[\begin{array}{ccc}1 & 0 & -2 \\0 & 1 & -1 \\0 & 0 & 0\end{array}\right]\left[\begin{array}{l}x_1 \\x_2 \\x_3\end{array}\right]=\left[\begin{array}{l}0 \\0 \\0\end{array}\right]$$[/tex]
x_3=t
[tex]$\Rightarrow \mathrm{x}_1=2 \mathrm{t}, \mathrm{x}_2=\mathrm{t}$[/tex]
[tex]$$\vec{x}=\left[\begin{array}{c}2 t \\t \\t\end{array}\right]=\left[\begin{array}{l}2 \\1 \\1\end{array}\right] t$$[/tex]
The Basis for the nullspace A-4 I will be [tex]\left(\left[\begin{array}{l}2 \\ 1 \\ 1\end{array}\right]\right)[/tex]
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What is the solution of 2x^2 5x 3 0?
Answers
The Given quadratic equation has two solutions: 1 and 3/2. A quadratic problem was solved by utilizing the factoring method.
A quadratic equation is defined.
At least one squared term must be present because a quadratic is a second-degree polynomial equation. It is also known as quadratic equations. ax^2 + b*x + c = 0 is the quadratic equation's general form.
where x is the unknown variable and a, b, and c are constant terms.
In this case, I'm solving a quadratic problem by factoring.
The following quadratic equation is
2x^2 - 5x + 3 = 0.
2x^2 - (3+2)x + 3 = 0
2x^2 -3x -2x + 3 = 0.
x(2x-3) -1(2x -3) = 0.
(2x-3)(x-1) = 0.
2x - 3 =0 or x -1 =0
x = 3/2 or x =1
So the solutions of a quadratic equation are 3/2 and 1.
Given Question is incomplete complete question here:
What are the solutions of the quadratic equation 2x^2-5x+3 = 0?
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The perimeter of triangle ABC is 26 units. The altitudes of triangle ABC are in the ratio 2: 3: 4. Compute the area of triangle ABC.
please help me w step by step explanation and work tysm !
Answers
Answer:
59
Step-by-step explanation:
I need some help with the problem from my math class (-4w - 7)5
Answers
-20w-35
I used the distributive property to distribute terms, and then simplified
The distance between two points A and B on latitude 60°S along their parallel of latitude is 2816km. (I) calculate the angular difference in their longitudes (ii) if A is due west of B and on longitude 20°W, find the longitude of B
Answers
Answer:
i) The angular difference in their longitude is approximately 50.65°
ii) The line of longitude where B is located is approximately 30.65°E
Step-by-step explanation:
The given parameters are;
The angle of latitude where the points A and B are located = 60°S
The distance between points A and B = 2,816 km
(i) The angular difference in their longitude, θ, is given as follows;
θ = 2,816/(2 × π × 6,371 × cos(60°)) × 360 ≈ 50.65°
The angular difference in their longitude, θ ≈ 50.65°
(ii) The location of the point A relative to B = Due west
The line of longitude on which A is located = 20°W
∴ The line of longitude where B is located, P = θ - 20°
∴ P = 50.65° - 20°W = 30.65°E
The line of longitude where B is located ≈ 30.65°E.
How do you write no solution?
Answers
No solution is when a system of equations has no solution. This means that the equations are inconsistent, and there is no set of values that can make the equations true.
For example, if the system of equations is 2x + 3y = 4 and 2x + 3y = 5, then there is no solution. This is because no matter what values are chosen for x and y, the equations cannot both be true. This means that the system of equations is inconsistent, and there is no solution.
To symbolically represent no solution, you can use the following notation:
2x + 3y = 4, 2x + 3y = 5 No solution
This notation is used to indicate that the equations are inconsistent and that there is no solution.
Complete question: What is the solution to the equation 2x + 3y = 4 and 2x + 3y = 5?
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A jetski races down the river at a constant speed for t seconds, covering m metres. How long, in
minutes, will it take for the jetski to cover n metres at the same speed?
Answers
It will take a time of n/60m minutes for the jetski to cover n metres at the same speed.
Define the term linear speed?The measurement of a moving object's actual distance travelled is called linear speed. Linear speed is the rate of motion of an object along a straight line.Thus, the angular speed times the r-dimensional distance equals the linear speed of such a point located on the object. The measuring unit is metres per second and metres per second. = the rate of change in radians per second.Speed = distance/ time
Speed = m/s
For the distance of n meters.
distance = speed * time
n = m/s*time
n = m /s* time
Time = n(m/sec)
Time = n/m * sec
Time = n/60m min
Thus, it will take n/60m minutes for the jetski to cover n metres at the same speed.
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PLEASE SOMEONE HELP MEEEE I AM LITERALLY FRYING NY BRAIN. NO LINKS OR I WILL REPORT YOU... PLEASE AND THANKS
Answers
Answer:
1. True
2. False
3. True
4. False
Step-by-step explanation:
What Is 8 1/8 - 4 3/8
Answers
Answer:
3 3/4 or 3 6/8
Step-by-step explanation:
hope this helps <3
9. The following illustration represents the height beam on the adult scale. Each small line represents 1/4 inch.
Each long line represents 1 inch. From the bottom to the break, readings are taken in an upward direction.
If a patient is taller than 50 inches, readings are taken in a downward direction and recorded directly at the
break in the scale.
Answers
Answer:
Helloooooooooo theree
Answer:
don't
Step-by-step explanation: it hard
The point P has coordinates (8, 9). In which quadrant does point P lie?
first quadrant
fourth quadrant
second quadrant
third quadrant
Answers
Answer:
The coordinates (8,9) lie in quadrant 1
Step-by-step explanation:
The coordinates on quadrant ___ go like this:
Quadrant l : (+,+)
Quadrant ll : (-,+)
Quadrant lll : (-,-)
Quadrant llll : (+,-)
Graph the image of square CDEF after a dilation with a scale factor of 2, centered at the
origin.
pls help lol
Answers
Answer:
Image attached
Step-by-step explanation:
Origin looks to be (0, -1), so move each point out by 2
multiple choice for final pls help me
Answers
Answer:
I hope this helps!
Step-by-step explanation:
ACB
Quadrilateral STUV is rotated 270° counterclockwise about the origin. If the coordinates of V were (3, -4), what are the coordinates of V’?
Answers
Given:
Quadrilateral STUV is rotated 270° counterclockwise about the origin.
The coordinates of V were (3, -4).
To find:
The coordinates of V’.
Solution:
If a figure rotated 270° counterclockwise about the origin, then the rule of rotation is:
[tex](x,y)\to (y,-x)[/tex]
The coordinates of V were (3, -4). Using the above rotation rule, we get
[tex]V(3,-4)\to V'(-4,-3)[/tex]
Therefore, the coordinates of V’ are (-4,-3).
A new water park was designed to contain two circular pools. Pool A has a diameter of 60 m, and pool B has a diameter of 80 m. a) If the designer wants to outline the pools with rope lights, how many meters will be needed to outline the smaller pool? (Use 3.14 for π.)
Answers
Answer:
140
Step-by-step explanation:
60+80=?......All you need to do is just add 60 and 80 and get the product
I hope that helped.
20
Mrs. Trailer has 1283.4 milliliters to divide
equally into 31 beakers. How many liters
will be in each beaker?
Answers
Answer:
about 5 milliliters
Step-by-step explanation:
IS THERE A SOLUTION ?? IF SO , WHAT IS IT ?? PLEASE HELP !! I WILL MARK BRAINLIEST !!!!
Answers
Answer:
y=4x+2
Step-by-step explanation:
-8x+2y=4
y=4x+2
−8x+2(4x+2)=4-8x+2(4x+2)=4
y=4x+2
−8x+8x+4=4-8x+8x+4=4
y=4x+2
0+4=40+4=4
y=4x+2
4=4
y=4x+2
What is the measure or a
Answers
The measures of the angles of the parallelogram are
The measure of ∠A = 72°
The measure of ∠B = 108°
What is a Parallelogram?A parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure
The four types are parallelograms, squares, rectangles, and rhombuses
Properties of Parallelogram
Opposite sides are parallel
Opposite sides are congruent
Opposite angles are congruent
Same-Side interior angles (consecutive angles) are supplementary
Each diagonal of a parallelogram separates it into two congruent triangles
The diagonals of a parallelogram bisect each other
Given data ,
Let the parallelogram be represented as ABCD
Now , the measure of ∠A = ( 5y - 3 )°
The measure of ∠C = ( 3y + 27 )°
For a parallelogram , opposite angles are congruent
So , measure of ∠A = measure of ∠C
Substituting the values in the equation , we get
( 5y - 3 )° = ( 3y + 27 )°
On simplifying the equation , we get
( 5y - 3 ) = ( 3y + 27 )
Subtracting 3y on both sides of the equation , we get
2y - 3 = 27
Adding 3 on both sides of the equation , we get
2y = 30
Divide by 2 on both sides of the equation , we get
y = 15
So , the measure of ∠A = ( 5y - 3 )°
The measure of ∠A = ( 5 ( 15 ) - 3 ) )°
The measure of ∠A = ( 75 - 3 )°
The measure of ∠A = 72°
Now , for a parallelogram , same-side interior angles (consecutive angles) are supplementary
So , the measure of ∠A + the measure of ∠B = 180°
Substituting the values in the equation , we get
72° + the measure of ∠B = 180°
Subtracting 72° on both sides of the equation , we get
The measure of ∠B = 108°
Hence , the measures of angles of the parallelogram are 72° and 108° respectively
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Time vs. Position of Battery Operated Car
5
4
Position (m)
2
0
5
10
Time (s)
15
20
What type of relationship is shown in the graph?
a non-linear, negative relationship
a linear, negative relationship
a linear, positive relationship
O a non-linear, positive relationship
Answers
A linear, negative relationship is the type of relationship which is shown in graph of Time vs. Position of Battery Operated Car .
what is linear equation ?A linear equation is an algebraic expression such as y=mx+b. The slope is m, and the y-intercept is B. A "linear equation in two variables" is the term used to describe the previous sentence, which has variables named y and x. Two different variables can be found in bivariate linear equations. Examples of linear equations include 2x - 3 = 0, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, and 3x - y + z = 3. The term "linear" refers to an equation having the form y=mx+b, where m stands for the slope and b for the y-intercept.
given
The area's 46-16 opposed to 14 is different from the area's 18 and T in a different way. The place is there. It isn't linearly getting farther apart. The lee is actually growing as a result.
Option B, the slanting upwards, is what we're going to go with. Since it will be linear, it isn't option C. It is untrue that the speaking has not been constant over the distance. Because it is a D-option, this increase-decrease is untrue. Option B is the best choice. Absolutely.
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Evie read an article that said 6\%6%6, percent of teenagers were vegetarians, but she thinks it's higher for students at her large school. To test her theory, Evie took a random sample of 252525 students at her school, and 20\%20%20, percent of them were vegetarians. To see how likely a sample like this was to happen by random chance alone, Evie performed a simulation. She simulated 404040 samples of n=25n=25n, equals, 25 students from a large population where 6\%6%6, percent of the students were vegetarian. She recorded the proportion of vegetarians in each sample. Here are the sample proportions from her 404040 samples: A dot plot for simulated sample proportions has a scale from 0.00 to 0.25 in increments of 0.01. The distribution is right skewed with dots plotted as follows. 0.00, 8. 0.004, 17. 0.08, 5. 0.12, 5. 0.16, 2. 0.20, 2. 0.24, 1. Evie wants to test H_0: p=6\%H 0 :p=6%H, start subscript, 0, end subscript, colon, p, equals, 6, percent vs. H_\text{a}: p>6\%H a :p>6%H, start subscript, start text, a, end text, end subscript, colon, p, is greater than, 6, percent where ppp is the true proportion of students who are vegetarian at her school. Based on these simulated results, what is the approximate ppp-value of the test? Note: The sample result was \hat p=20\% p ^ =20%p, with, hat, on top, equals, 20, percent. Choose 1 answer: Choose 1 answer: (Choice A) A p\text{-value}\approx0.01p-value≈0.01p, start text, negative, v, a, l, u, e, end text, approximately equals, 0, point, 01 (Choice B) B p\text{-value}\approx0.025p-value≈0.025p, start text, negative, v, a, l, u, e, end text, approximately equals, 0, point, 025 (Choice C) C p\text{-value}\approx0.03p-value≈0.03p, start text, negative, v, a, l, u, e, end text, approximately equals, 0, point, 03 (Choice D) D p\text{-value}\approx0.075p-value≈0.075
Answers
Answer:
p-value~ 0.075
Step-by-step explanation:
( Choice D on Khan Academy)
The closest answer choice is (Choice A) A p-value ≈ 0.01, is the best approximation among the answer choices given.
What is p-value?In a statistical hypothesis test, the p-value is a probability value that measures the strength of evidence against the null hypothesis.
It is the likelihood of observing a test statistic that is as extreme as or more extreme than the observed test statistic, assuming the null hypothesis is true.
To find the approximate p-value of the test, we need to calculate the proportion of simulated sample proportions that are greater than or equal to the sample proportion of 20%.
From the dot plot, we can see that only a few simulated sample proportions are greater than or equal to 20%, which suggests that the p-value is small.
Counting the dots from the dot plot, we can see that there are a total of 8 + 17 + 5 + 5 + 2 + 2 + 1 = 40 simulated sample proportions that are greater than or equal to 20%.
Therefore, the proportion of simulated sample proportions that are greater than or equal to 20% is 40/40 = 1.
Since the p-value is the probability of observing a sample proportion as extreme or more extreme than the observed sample proportion, given that the null hypothesis is true, we can approximate the p-value as the proportion of simulated sample proportions that are greater than or equal to the observed sample proportion.
Therefore, the approximate p-value is 1, which is very small and suggests strong evidence against the null hypothesis.
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Given the matrices, calculate AB
Answers
Answer:
C11=-2
C22=8
hope this will help you more
good luck
Composite Figures Question 9.
A. 57 ft
B. 288 ft
C. 208 ft
D. 264 ft
Answers
Answer:
D
Step-by-step explanation:
find area of first rectangle
18 x 6 = 108
find area of second rectangle
20 x 5 = 100
find area of two triangles
7 x 8 =56
56/2 = 28
each triangle has an area of 28
28+28+100+108 = 264ft
You have four sweaters, five pairs of pants, and three pairs of shoes. How many different combinations can you make, wearing one of each
Answers
Therefore, 60 different combinations can be made wearing one of each type
What is Fundamental Counting Principle? If an event can occur in m different ways, and another event can occur in n different ways, then the total number of occurrences of the events is m × n.This video uses manipulatives to review the five counting principles including stable order, correspondence, cardinality, abstraction, and order irrelevance. When students master the verbal counting sequence they display an understanding of the stable order of numbers.Fundamental Principle of Counting helps to determine how selection is done on the basis of available choices. Fundamental Principle of Counting simplifies the approach of selection considering all the possible choices and their combinations for calculating the ProbabilityThe fundamental counting principle states that if there are p ways to do one thing, and q ways to do another thing, then there are p×q ways to do both things. Example 1: Suppose you have 3 shirts (call them A , B , and C ), and 4 pairs of pants (call them w , x , y , and z ). Then you have. 3×4=12.To learn more about Fundamental Counting Principle refers to:
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Can u help me with this
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Answer:
D) (6.6, 7.4)
Step-by-step explanation:
Begin by finding the margin of error based on the given information.
[tex]\boxed{\begin{minipage}{7.3cm}\underline{Margin of Error formula}\\\\$\textsf{Margin of Error}= Z \times \dfrac{S}{\sqrt{n}}$\\\\where:\\\phantom{ww} $\bullet$ $Z =$ $Z$ score\\\phantom{ww} $\bullet$ $S =$ Standard Deviation of a population\\\phantom{ww} $\bullet$ $n =$ Sample Size\\\end{minipage}}[/tex]
Given:
S = 1.3n = 50The z-score for 95% confidence level is 1.96.
Therefore, to find the margin of error, substitute the values into the formula:
[tex]\implies \textsf{Margin of Error}= 1.96 \times \dfrac{1.3}{\sqrt{50}}[/tex]
[tex]\implies \textsf{Margin of Error}=0.360341...[/tex]
To find a reasonable range for the true mean number of hours a teenage spend on their phone, subtract and add the margin of error to the given mean of 7 hours:
[tex]\begin{aligned}\implies \textsf{Lower bound}&=7-0.360341...\\&=6.63965...\\&=6.6\;\; \sf (1\;d.p.)\end{aligned}[/tex]
[tex]\begin{aligned}\implies \textsf{Upper bound}&=7+0.360341...\\&=7.360341...\\&=7.4\;\; \sf (1\;d.p.)\end{aligned}[/tex]
Therefore, the reasonable range for the true mean is:
(6.6, 7.4)One computer in a lab is programmed to back up data at the turn of the minute every five minutes. Another computer is programmed to back up data at the turn of the minute every two minutes. Find the number of times in twenty-four hours that the two computers back up data at the same time.
(Assume that the computers do not back up at the start of the 24-hour period.)
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The number of times which the two computers back up at the same time in twenty four hours as required is; 144.
How many times does the computers back up at the same time in 24 hours?It follows from the task content that the first computer backs up after every five minutes and the second computer backs up every two minutes.
On this note, the least common multiple of both backup times represents when the two computers back up at the same time.
Hence, since the LCM of 5 and 2 is; 10.
It follows that the two computers back up at the same time after every ten minutes.
Ultimately, the number of times for which they backup at the same time in 24 hours; since 60 minutes = 1 hour is;
= ( 24 × 60 ) / 10
= 24 × 6
= 144 times.
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Step-by-step explanation:
[tex]2\pi {r}^{2} + 2\pi rh[/tex]
[tex]2(3.14) {(20)}^{2} + 2(3.14)(20)(11)[/tex]
[tex]6.28(400) + 6.28(220) \\ 2512 + 1381.6 = 3893.6 {mi}^{2} [/tex]