Answer:
[tex]\bar x = 107.11[/tex]
[tex]\sigma_x = 31.07[/tex]
Step-by-step explanation:
See comment for complete question
Given
[tex]x: 97\ 178\ 129\ 90\ 75\ 94\ 116\ 100\ 85[/tex]
Solving (a): The sample mean
This is calculated using:
[tex]\bar x = \frac{\sum x}{n}[/tex]
So, we have:
[tex]\bar x = \frac{97+ 178+ 129+ 90+ 75+ 94+ 116+ 100+ 85}{9}[/tex]
[tex]\bar x = \frac{964}{9}[/tex]
[tex]\bar x = 107.11[/tex]
Solving (b): The sample standard deviation
This is calculated as:
[tex]\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}[/tex]
So, we have:
[tex]\sigma_x = \sqrt{\frac{(97 - 107.11)^2 +.............+ (85- 107.11)^2 }{9-1}}[/tex]
[tex]\sigma_x = \sqrt{\frac{7720.8889}{8}}[/tex]
[tex]\sigma_x = \sqrt{965.1111125}[/tex]
[tex]\sigma_x = 31.07[/tex]
Hector's Position:
Hector was standing halfway between first and second base, at the grass line. The
grass line is 95 feet from the pitcher's mound.
6. Calculate the coordinates for Hector's position. [Note: We can assume that 95
feet is an approximately horizontal distance from the pitcher's mound to the grass
line.] (2 points: 1 for x, 1 for y)
Hector was standing at the coordinate ( __, _).
Calculate Hector's Throw:
Answer:
(137.78, 47.72)
Step-by-step explanation:
(I just finished this assignment.)
Tre's position at the pitcher's mound as the point (42.78, 42.78).
( x , y )
Hector is about 95 feet away from the pitcher's mound horizontal, (x axis).
Since we already have the correct y-coordinate, we need to solve for the correct x-coordinate.
x = 95 + 42.78
↓ ↓ ↓
95 + 42.78 = 137.72
Now all you need to do is write out the coordinates.
Hector's coordinates are (137.72, 47.78 )
An educator wants to see how the number of absences for a student in her class affects the student’s final grade. The data obtained from a sample are shown.
Number of absences
26
29
32
34
36
37
Final grade
48
68
66
69
76
67
The Question is solved with a simple linear equation and is explained as follows:
Simple linear equation is :
Y = 86.784 - 2.67x
The educator wants to see the relationship between no of absences and final grades of the student.
No. of absence of student is independent variable while final grade is dependent variable.
The final grades of the student are dependent on the no. of absence they do.
b = Sxy / Sxx
b = { 2312 [ 37 * 422 ] / 6 } / 337 - 37^2 / 6 = -2.67
a = -2.67 * 37/6 = 86.784
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Heeeelp pleasssse :D
Answer:
(x - 3/8)^2 = x^2 - 3/4x + 9/64
Step-by-step explanation:
Step-by-step explanation:
divide the number with x by 2 and get the square of that number and add that number to this given equation
number with x = -3/4
= x^2 - 3/4 x + 9/64
= (x -3/8) ^2
The owner of a small hardware store employs three men and three women. He will select three employees at random to work on Christmas Eve. Find the probability that exactly two men (and one woman) will be selected to
work (2DP)
Answer:
Step-by-step explanation:
This is a hypergeometric distribution
[tex]\frac{{3\choose2}*{3\choose1}}{{6\choose3}}[/tex]=.45
power sharing helps the ruling party to retain power for a long time. tick or wrong
Charlie has an annual salary of $75,000.00. He is paid every two weeks. What is the gross income amount for each paycheck?
Answer:
$2884.62
Step-by-step explanation:
A year has 52 weeks
The number of times Charlie will receive a paycheck will be 52w ÷ 2w = 26 times
Charlie's gross income each paycheck will be 7500÷26 = $2884.62 every two weeks
75000 ÷ (52 ÷2)
7500 ÷ 26
$2884.62
Select the correct answer.
What is the best way to describe a theme of this poem?
A.
The main purpose of having New Year's resolutions is to make people feel bad.
B.
The failures of the past should inspire people to accomplish more in the future.
OC.
By the end of the year, it is too late to make any changes to a person's life.
D.
People would accomplish their New Year's resolutions if they wrote them down.
B.The failures of the past should inspire people to accomplish more in the future.
write the expression as a decimal , 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000 =__
Answer:
6.986.
Step-by-step explanation:
6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000
We do the multiplications first ( according to PEMDAS):-
= 6 + 9 * 0.1 + 8 * 0.01 + 6 * 0.001
= 6 + 0.9 + 0.08 + 0006
= 6.9 + 0.086
= 6 986.
The value of the equation in the decimal form is A = 6.986
What is an Equation?
Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.
It demonstrates the equality of the relationship between the expressions printed on the left and right sides.
Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.
Given data ,
Let the equation be represented as A
Now , the value of A is
A = 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000
On simplifying the equation , we get
The value of 6 x 1 = 6
The value of 9 x 1/10 = 0.9
The value of 9 x 1/100 = 0.08
The value of 6 x 1/1000 = 0.006
So , substituting the values in the equation A , we get
A = 6 + 0.9 + 0.08 + 0.006
On simplifying the equation , we get
A = 6.986
Therefore , the value of A is 6.986
Hence , the value of the equation is 6.986
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How many ways can a president, vice president, secretary, and treasurer be chosen from a club with 8 member
Answer:
504
Step-by-step explanation:
Diện tích xung quanh của hình chóp tứ giác đều có cạnh bằng 6cm và độ dài trung đoạn bằng 10cm là:
A. 120 cm2 B. 240 cm2 C. 180 cm2 D. 60 cm2
Answer:
B. 240 cm2
Step-by-step explanation:
Chu vi đáy: 10x=40
Diện tích xung quanh: Sxq=1/2 x40x12=240
A triangular lamina has vertices (0, 0), (0, 1) and (c, 0) for some positive constant c. Assuming constant mass density, show that the y-coordinate of the center of mass of the lamina is independent of the constant c.
The equation of the line through (0, 1) and (c, 0) is
y - 0 = (0 - 1)/(c - 0) (x - c) ==> y = 1 - x/c
Let L denote the given lamina,
L = {(x, y) : 0 ≤ x ≤ c and 0 ≤ y ≤ 1 - x/c}
Then the center of mass of L is the point [tex](\bar x,\bar y)[/tex] with coordinates given by
[tex]\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m[/tex]
where [tex]M_x[/tex] is the first moment of L about the x-axis, [tex]M_y[/tex] is the first moment about the y-axis, and m is the mass of L. We only care about the y-coordinate, of course.
Let ρ be the mass density of L. Then L has a mass of
[tex]\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2[/tex]
Now we compute the first moment about the y-axis:
[tex]\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6[/tex]
Then
[tex]\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3[/tex]
but this clearly isn't independent of c ...
Maybe the x-coordinate was intended? Because we would have had
[tex]\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6[/tex]
and we get
[tex]\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13[/tex]
The center of mass for a uniform triangular shape is on its centroid. The y-coordinate of the center of mass of the lamina is 1/3 (independent of c).
What is the center of mass for a triangular shape?If the surface is plane triangle approximately and mass is uniformally distributed, then its center of mass will lie on the centroid of that triangle.
What is centroid of a triangle and its coordinates?The point of intersection of a triangle's medians is its centroid (the lines joining each vertex with the midpoint of the opposite side).
If the triangle has its vertices as [tex](x_1, y_1), (x_2, y_2) , \: (x_3, y_3)[/tex], then the coordinates of the centroid of that triangle is given by:
[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)[/tex]
For this case, the triangular lamina has vertices (0, 0), (0, 1) and (c, 0)
Assuming its mass is spread regularly, the coordinates of its center of mass would be:
[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)\\\\(x,y) = \left( \dfrac{0+0+c}{3} + \dfrac{0+1+0}{3} \right) = (c/3, 1/3)[/tex]
Thus, the y-coordinate of the center of mass of the lamina is 1/3 (independent of c).
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Help please guys thanks
Answer:
D
Step-by-step explanation:
sqrt_{4}(81)^5=(81^(5))^(1/4)=81^(5/4)
Answer:
D
Step-by-step explanation:
if it was properly typed, it would have been All of the above but the most correct option is D.
Here is a number sequence. The rule for finding the next term is to add
a, where a is an integer. ! ! 8 ........! ! ........! ! 29 Work out the two
missing terms.
Answer:
8,15,22,29
Step-by-step explanation:
the interger a is 7,so to find the next term you have to add 7 plus the 8,
8+7=15
15+7=22
22+7=29
8,15,22,29
I hope this helps
write your answer as an integer or as a decimal rounded to the nearest tenth
Answer:
123456-6-&55674
Step-by-step explanation:
rdcfvvzxv.
dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see
recall see
8/11 - Applying the Distributive Property
1. What is the simplest expression equivalent to 5(2x - 13) ?
HELP HELP! I NEED URGENT HELP WITH THIS equashin.
Answer:
V = 1071.79 yd^3
Step-by-step explanation:
The volume of a cone is
V = 1/3 pi r^2 h where r is the radius and h is the height
We are given a diameter of 16 so the radius is 1/2 of the diameter or 8
The height is 16
V = 1/3 ( 3.14) (8)^2 ( 16)
V = 1071.78666 yd^3
Rounding to the nearest hundredth
V = 1071.79 yd^3
[tex] \large\begin{gathered} {\underline{\boxed{ \rm {\red{Volume \: of \: Cone \: = \: \pi \: {r}^{2} \: \frac{h}{3} }}}}}\end{gathered}[/tex]
[tex] \bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{Diameter}{2} \\ [/tex]
[tex]\bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{16 \: yd}{2} \\ [/tex]
[tex]\bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{ \cancel{16 \: yd} \: \: ^{8} }{ \cancel{2}} \\ [/tex]
[tex]\bf{\red{ \longrightarrow}} \tt \: \large{\bf{{{\color{navy}{r \: = \: 8 \: yd}}}}}[/tex]
[tex]\bf{\red{ \longrightarrow}} \tt \: \: \large{\bf{{{\color{navy}{h \: = \: 16 \: yd}}}}}[/tex]
[tex] \bf \large \longrightarrow \: \: 3.14 \: \times \: {8}^{2} \: \times \: \frac{16}{3} \\ [/tex]
[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: \frac{16}{3} \\ [/tex]
[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: \frac{ \cancel{16} \: \: ^{5.33} }{ \cancel{3}} \\ [/tex]
[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: 5.33[/tex]
[tex]\bf \large \longrightarrow \: \:200.96 \: \times \: 5.3[/tex]
[tex]\bf \large \longrightarrow \: \:1071.79[/tex]
Option (A) is the correct answer
Ann,Ryan and Keith have a total of $114 in their wallets. Ryan has three times what Keith has. Keith has nine dollars less than Ann. how much do they have in their wallets?
Answer:
Ann has $30
Ryan has $63
Keith has $21
Step-by-step explanation:
First I assigned variables to the amounts each owns. It doesn't really matter what letters, but I said x represents Ann's money, y represents Ryan's money, and z represents Keith's money.
We know that with all their money added up, they have 114. This gives us our first equation;
114=x+y+z
Ryan has 3 times what Keith has, which, in terms of variables, is
y=3z
Keith has nine dollars less than Ann. Less, in word problems, always means subtraction. This gives us our third equation:
z=x-9
Okay so the impulse is to put z=x-9 and y=3z into the first equation, but the goal is to create an equation with only one variable so this will not work.
What we do instead is sub z=x-9 into y=3z. This gives
y=3(x-9)=3x-27
Now that both y and z are in terms of x, we plug those equations into the first and then solve for x;
114=x+3x-27+x-9
114=5x-36
150=5x
30=x
Now that we know x, we can find z and y using the equations from earlier:
z=30-9=21
y=3(21)=63
x=30, y=63, and z=21, thus Ann has $30, Ryan has $63, and Keith has $21.
What type of line is PQ?
A. altitude
B. angle bisector
C. side bisector
D. median
The line PQ of the triangle is an altitude. The correct option is A.
What is the altitude of the triangle?
A line segment passing through a triangle's vertex and running perpendicular to the line containing the base is the triangle's height in geometry.
The extended base of the altitude is the name given to this line that contains the opposing side. The foot of the altitude is the point at where, the extended base and the height converge.
In the given triangle the line segment PQ is passing through a triangle's vertex and running perpendicular to the line containing the base is the triangle's height in geometry.
Therefore, the line PQ of the triangle is an altitude. The correct option is A.
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Which is the solution to-x/2<-4
A x<-8
B x2-8
C x <8
D x 8
Answer:
A.x<-8
Step-by-step explanation:
=1/2x<−4
=2*(1/2x)< (2)*(-4)
= x<-8
f(t)= 102,000/1+4400e^-t
Answer:
Beginning (t=0) population with flu is 23.
After 4 weeks, population with flu is 1250.
After an infinite amount of weeks, the population witf flu is 102000
Step-by-step explanation:
First question asks you to replace t with 0 because it says beginning.
102000/(1+4400e^-0)=102000/(1+4400)=102000/4401=23.17655 approximately. To nearest whole number this is 23.
After 4 weeks means we replace t with 4:
102000/(1+4400e^-4)
Calculator time:
1250.17142 which to nearest whole number is 1250
If t is super large, then e^-t is super close to 0.
So the limiting number is
102000/(1+4400×0)=102000/1=102000
Instructions: State what additional information is required in order
to know that the triangles in the image below are congruent for the
reason given
Reasory. ASA Postulate
Answer:
ACB = JCB
Step-by-step explanation:
ASA means angle - (included) side - angle.
we have one angle confirmed.
we have the connected side BC confirmed (the diagram shows that the side is shared, so it is not only congruent, it is actually identical).
now we need confirmation for the angle at the other end point of that side.
2/8 of a rope is 28 meters.What is the length of the rope? A.32 B.42 C.4 D.21
let length be x
ATQ
[tex]\\ \sf\longmapsto \dfrac{2}{8}\times x=28[/tex]
[tex]\\ \sf\longmapsto \dfrac{2x}{8}=28[/tex]
[tex]\\ \sf\longmapsto \dfrac{x}{4}=28[/tex]
[tex]\\ \sf\longmapsto x=4(28)[/tex]
[tex]\\ \sf\longmapsto x=112[/tex]
Step-by-step explanation:
there is something wrong with your problem description.
the offered answer options do not fit to the solution as it is described.
2/8th of a rope is 28 meters long. how long is the whole rope ?
as the other answer said : 2/8 = 1/4
and 1/4th of the rope x = 28 m
1/4 × x = 28
x (the length of the whole rope) = 4×28 = 112 meters
but - maybe the original problem said that 7/8th (and not 2/8th) of a rope is 28 m.
7/8 × x = 28
1/8 × x = 4
x = 32 m
then A (32) would be the right answer !
d is none of the above , and yes
Answer:
[tex] = 2 {}^{2} - 3(2) = - 2 \\ 3 {}^{2} - 3(3) = 0 \\ 4 {}^{2} - 3(4) = 4 \\ 5 {}^{2} - 3(5) = 10[/tex]
A car travels at a constant speed towards a town. If it increases its speed by 15 km/h, the time required is in a ratio of 6 : 5. If it reduces its speed by 15 km/h, it needs another 105 minutes to arrive at the destination. Find the distance travelled by the car.
9514 1404 393
Answer:
525 km
Step-by-step explanation:
Let d represent the distance to the town. Let s represent the nominal speed of the car. The relation between time, speed, and distance is d = st.
t1 = d/s
t2 = d/(s+15)
t1 : t2 = 6 : 5 . . . increasing the speed reduces the time
Substituting for t1 and t2, we have ...
(d/s)/(d/(s+15)) = 6/5
(s +15)/s = 6/5
1 +15/s = 1 +1/5
s = 5·15 = 75 . . . . nominal speed in km/h
__
Decreasing the speed increases the time.
d/75 +(105/60) = d/(75-15)
d(60/75) +105 = d . . . . . . multiply by 60
105 = d/5 . . . . . . . . . . . subtract 4/5d
525 = d . . . . . . . . . . multiply by 5
The distance traveled by the car is 525 km.
240) What term is 359 in the sequence 5, 12, 17, 23, 28, 29......?
Answer:
Check your question again
Step-by-step explanation:
The arithmetic equation of this sequence is an=5+(n-1)*7. Replace 359 with an and solve for n
359=5+(n-1)*7, 354/7=n-1. Wait you got the whole equation wrong, the first term should be 7 so that the common difference be equal to 5
To calculate the volume of a chemical produced in a day a chemical manufacturing company uses the following formula below:
[tex]V(x)=[C_1(x)+C_2(x)](H(x))[/tex]
where represents the number of units produced. This means two chemicals are added together to make a new chemical and the resulting chemical is multiplied by the expression for the holding container with respect to the number of units produced. The equations for the two chemicals added together with respect to the number of unit produced are given below:
[tex]C_1(x)=\frac{x}{x+1} , C_2(x)=\frac{2}{x-3}[/tex]
The equation for the holding container with respect to the number of unit produced is given below:
[tex]H(x)=\frac{x^3-9x}{x}[/tex]
a. What rational expression do you get when you combine the two chemicals?
b. What is the simplified equation of ?
c. What would the volume be if 50, 100, or 1000 units are produced in a day?
d. The company needs a volume of 3000 How many units would need to be produced in a day?
Answer:
[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
[tex]V(50) = 2548.17[/tex] [tex]V(100) = 10098.10[/tex] [tex]V(1000) = 999201.78[/tex]
[tex]x = 54.78[/tex]
Step-by-step explanation:
Given
[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]
[tex]C_1(x) = \frac{x}{x+1}[/tex]
[tex]C_1(x) = \frac{2}{x-3}[/tex]
[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]
Solving (a): Expression for V(x)
We have:
[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]
Substitute known values
[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]
Solving (b): Simplify V(x)
We have:
[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]
Solve the expression in bracket
[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]
[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]
Factor out x
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]
Express as difference of two squares
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]
Cancel out x - 3
[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
Solving (c): V(50), V(100), V(1000)
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
Substitute 50 for x
[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]
[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]
[tex]V(50) = 2548.17[/tex]
Substitute 100 for x
[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]
[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]
[tex]V(100) = 10098.10[/tex]
Substitute 1000 for x
[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]
[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]
[tex]V(1000) = 999201.78[/tex]
Solving (d): V(x) = 3000, find x
[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]
Cross multiply
[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]
Equate to 0
[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]
Open brackets
[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]
Collect like terms
[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]
[tex]x^3 + x^2 -3001x -2994 = 0[/tex]
Solve using graphs (see attachment)
[tex]x = -54.783[/tex] or
[tex]x = -0.998[/tex] or
[tex]x = 54.78[/tex]
x can't be negative. So:
[tex]x = 54.78[/tex]
The management of a large airline wants to estimate the average time after takeoff taken before the crew begins serving snacks and beverages on their flights. Assuming that management has easy access to all of the information that would be required to select flights by each proposed method, which of the following would be reasonable methods of stratified sampling?
a. For each day of the week, randomly select 5% of all flights that depart on that day of the week.
b. Divide all flights into the following 4 groups on the basis of scheduled departure time: before 9:00 am 9:00 am to 1:00 pm. 1:00 pm to 5:00 pm, and after 5:00 pm. Randomly select 5% of the flights in each group.
c. For each crew member the airline employs, randomly select 5 flights that the crew member works.
d. Divide the airports from which the airline's fights depart into 4 regions: Northeast, Northwest Southwest and Southeast. Randomly select 5% of all flights departing from airports in each region
Answer:
The answer is "Option a, Option b, and Option d".
Step-by-step explanation:
In the given question it is used to stratifying the sampling if the population of this scenario it flights takes off when it is divided via some strata.
In option a, In this case, it stratified the sampling, as the population of planes taking off has been divided into the days of the week. In option b, It also used as the case of stratified sampling. In options c, it is systematic sampling, that's why it is wrong. In option d, It is an example of stratifying the sampling.Answer:
For each day of the week, randomly select 5% of all flights that depart on that day of the week.
Divide all flights into the following 4 groups on the basis of scheduled departure time: before 9:00 am, 9:00 am to 1:00 pm, 1:00 pm to 5:00 pm, and after 5:00 pm. Randomly select 5% of the flights in each group.
Divide the airports from which the airline's flights depart into 4 regions: Northeast, Northwest, Southwest, and Southeast. Randomly select 5% of all flights departing from airports in each region.
Step-by-step explanation:
ll sampling methods that divide the flights into a small number of mutually exclusive categories are appropriate. These methods include all flights on the basis of a characteristic that might be associated with the variable being investigated and randomly selects a proportionate number of flights from each group.
A ball is dropped from a height of 14 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen. (a) Find the total distance the ball has traveled at the instant it hits the ground the fourth time. (Enter an exact number.)
Answer:
Hello,
742/27 (ft)
Step-by-step explanation:
[tex]h_1=14\\\\h_2=\dfrac{14}{3} \\\\h_3=\dfrac{14}{9} \\\\h_4=\dfrac{14}{27} \\\\[/tex]
[tex]d=14+2*\dfrac{14}{3} +2*\dfrac{14}{9} +2*\dfrac{14}{27} \\=14*(1+\dfrac{1}{3}+\dfrac{2}{9} +\dfrac{2}{27} )\\=14*\dfrac{53}{27} \\=\dfrac{742}{27} \\[/tex]
The total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]
What is the total distance?
Distance is a numerical measurement of how far apart objects or points are. It is the actual length of the path travelled from one point to another.
Here given that,
A ball is dropped from a height of [tex]14[/tex] ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.
So, after striking with the ground it covers the distance [tex]14[/tex] ft. so it rebounds to the height is [tex]\frac{1}{3}(14)[/tex].
Then again it hits the ground and covers the distance [tex]\frac{1}{3}(14)[/tex] and again after rebounding it goes to the height is
[tex]\frac{1(1)}{3(3)}.(14)=\frac{(1)^2}{(3)^2}(14)[/tex]
Then it falls the same distance and goes back to the height
[tex]\frac{1}{3}[/tex] ×[tex](\frac{(1)^2}{(3)^2})[/tex] ×[tex]14[/tex] = [tex]\frac{(1)^3}{(3)^3}(14)[/tex]
So, the total distance travelled is
[tex]14+2[\frac{1}{3}(14)+(\frac{1}{3})^2(14)+(\frac{1}{3})^3(14)+...][/tex]
We take the sum is twice because it goes back to the particular height and falls to the same distance.
[tex]S=14+2(\frac{\frac{1}{3}(14)}{1-\frac{1}{3}})\\\\\\S=\frac{a}{1-r}\\\\\\S=14+2(\frac{\frac{14}{3}}{\frac{2}{3}})\\\\S=14+2(\frac{14}{2})\\\\S=14+2(7)\\\\S=14+14\\\\S=28ft[/tex]
Hence, the total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]
To know more about thetotal distance
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find the value and express it in standard form : 5×10^8×2×10^11 please answer
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guys help me I really need your help
Answer:
a x^2/2 is a polynomial because the power of x is 2 which is a positive whole number but 2/x^2 is not a polynomial because the power of x is -2 which is negative whole number.
b.in
[tex] \sqrt{2 x} [/tex]
the power if x will be
[tex]x {}^{ \frac{1}{2} } [/tex]
which is not a whole number so it is not a polynomial.
but in
[tex] \sqrt{2} x[/tex]
the power if x is a positive whole number.so it is a polynomial.
c.the greatest power of variable of the term is called degree of polynomial
