a) y = 3/5x + 11/5
b) y = -5/3x + 9
Explanation:[tex]\begin{gathered} a)\text{ }y\text{ = }\frac{3}{5}x\text{ - 3} \\ \text{compare with equation of line:} \\ y\text{ = mx + b} \\ m\text{ =slope, b = y-intercept} \\ m\text{ =slope = 3/5} \\ b\text{ = -3} \end{gathered}[/tex]For a line to be parallel to another line. the slope of the 1st line will be equalt to the slope of the 2nd line:
slope of 1st line = 3/5
So, the slope of the 2nd line = 3/5
Given point: (3, 4) = (x, y)
To get the y-intercept of the second line, we would insert the slope and the point into the equation of line
[tex]\begin{gathered} y\text{ = mx + b} \\ 4\text{ = }\frac{3}{5}(3)\text{ + b} \\ 4\text{ = 9/5 + b} \\ 4\text{ - }\frac{\text{9}}{5}\text{ = b} \\ \frac{20-9}{5}\text{ = b} \\ b\text{ = 11/5} \end{gathered}[/tex]The equation of line parallel to y = 3/5x - 3:
[tex]\begin{gathered} y\text{ = mx + b} \\ y\text{ = }\frac{3}{5}x\text{ + }\frac{11}{5} \end{gathered}[/tex][tex]b)\text{ line perpendicular to y = 3/5x - 3}[/tex]For a line to be perpendicular to another line, the slope of one will be the negative reciprocal of the second line
Slope of the 1st line = 3/5
reciprocal of 3/5 = 5/3
negative reciprocal = -5/3
slope of the 2nd line (perpendicular) = -5/3
We need to get the y-intercept of the perpendicular line:
[tex]\begin{gathered} \text{given point: (3,4) = (x, y)} \\ y\text{ = mx + b} \\ m\text{ of the perpendicular = -5/3} \\ 4\text{ = }\frac{-5}{3}(3)\text{ + b} \\ 4\text{ = -5 + b} \\ 4\text{ + 5 = b} \\ b\text{ = 9} \end{gathered}[/tex]The equation of line perpendicular to y = 3/5x - 3:
[tex]\begin{gathered} y\text{ = mx + b} \\ y\text{ = }\frac{-5}{3}x\text{ + 9} \end{gathered}[/tex]Make an estimate. Then divide using partial-quotients division. Write your remainder as a fraction and pls make it make sense
Notice that 812 is close to 810 and 17 is closer to 15 than to 20; thus, a possible estimate is
[tex]\frac{810}{15}=54[/tex]However, 800/20 is a more straightforward approximation. Both can be used since you are not being asked for a specific approximation.
Partial-quotients division
Thus, the answer is quotient equal to 47 and remainder equal to 13/17.
!!PLEASE ANSWER FAST PLEASE!! Given f(x)=(1/4)(5-x)² what is the value of f(11)
Answer:
f(11) = 9
Explanation:
The equation for f(x) is:
[tex]f(x)=\frac{1}{4}(5-x)^2[/tex]To know the value of f(11), we need to replace x by 11 and solve, so:
[tex]\begin{gathered} f(11)=\frac{1}{4}(5-11)^2 \\ f(11)=\frac{1}{4}(-6)^2 \\ f(11)=\frac{1}{4}(36) \\ f(11)=9 \end{gathered}[/tex]Therefore, the value of f(11) is 9.
used to figure for exercises two through nine period determine whether each pair of lines are parallel or perpendicular. right yes or no
2. q and v are parallel. YES
3. r and s are parallel. NO
4. r and t are parallel. NO
5. s and u are parallel. YES
6. q and s are perpendicular. NO
7. q and v are perpendicular. NO
8. r and s are perpendicular. YES
9. t and v are perpendicular. YES
Which number line represents the solution to the inequality
–4x – 12 < 12 ?
PLEASE ANSWER FAST
Answer:
x ≥ -6
Option C
Step-by-step explanation:
Hello!
We can solve the inequality by isolating x. Remember, flip the sign when you divide or multiply both sides by a negative number in an inequality.
Solve for x-4x - 12 ≤ 12-4x - 12 + 12 ≤ 12 + 12-4x ≤ 24-4x / -4 ≥ 24 / -4 => Flip the sign!x ≥ -6The answer is option c, all values greater than -6.
For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 35 N acts on a certain object, the acceleration of the object is 5 m/s^2. If the force is changed to 49 N what will be the acceleration of the object?
Answer:The acceleration that an objects gains is given by the mass of the object.
If the acceleration of the object becomes 5 m/s² the force is 15 N.
Reason:
The given parameters are;
The acting force ∝ The acceleration of the object.
The acceleration given by an amount of force, F, of 18 N = 6 m/s²
Required:
The force acting on the object acceleration, a, is 5 m/s².
Solution:
According to Newton's Second Law of motion, we have;
F = m·a
Where;
m = The mass of the object
Therefore, we have;
From the conditions, F = 18 N, when a = 6 m/s², we have, the mass of the
given object is given as follows;
The force acting when the the acceleration, a = 5 m/s², is therefore;
F = 3 kg × 5 m/s² = 15 N
If the acceleration of the object becomes 5 m/s² the force is 15 N.
Step-by-step explanation:
In the figure, what set of angles are corresponding angles?angle 6 and angle 7angle 5 and angle 7angle 1 and angle 7angle 4 and angle 7
Step 1
In geometry, corresponding angles are formed where a line known as an intersecting transversal, crosses through a pair of straight lines. Corresponding angles are the pairs of angles that are found in the same relative position on different intersections.
Step 2
Find the set of corresponding angles
[tex]\text{Angle 4 and angle 7}[/tex]Hence, the set of corresponding angles is; angle 4 and angle 7
Theo sales person makes $350 each week plus an additional $28 per sale. Theo wants his paycheck to be at least $550 each week. Solve the inequality and choose the best answer to the scenario.
I need help with this assignment
In this problem the big angle will be equal to two times the smaller angle so the correct expression is:
[tex]\angle TUV=2\angle VUW[/tex]Now we can rewrite the equation so:
[tex]\angle VUW=\frac{1}{2}\angle TUV[/tex]I need help solving this I’m having trouble with it is from my trigonometry prep bookIf you can **** use Desmos to graph the function that is provided in the picture
So we have to graph the function:
[tex]f(x)=\cot (x+\frac{\pi}{6})[/tex]First is important to note that the cotangent can be defined by the quotient between the cosine and the sine:
[tex]\cot (x+\frac{\pi}{6})=\frac{\cos(x+\frac{\pi}{6})}{\sin(x+\frac{\pi}{6})}[/tex]By looking at this new expression we can infer a few things about the graph. First of all, we have a sine in the denominator which means that the denominator can be equal to 0. Let's assume that the denominator is 0 at x=a. Then the graph has a vertical asymptote at x=a. What's more, the sine is a periodic funtion that is equal to zero for an infinite amount of x values so the graph of the cotangent has infinite vertical asymptotes. The good part is that we just need to graph one full period and in the case of the cotangent one full period is completed between two consecutive vertical asymptote. So basically we have to find two consecutive vertical asymptote and graph the function between them.
So let's begin by finding two x values that makes the denominator equal to 0. The sine is equal to 0 when its argument is equal to 0 and the next value at which the sine is equal to zero is pi so:
[tex]\sin 0=0=\sin \pi[/tex]Then we can construct two equations:
[tex]\begin{gathered} \sin (x+\frac{\pi}{6})=0=\sin 0 \\ \sin (x+\frac{\pi}{6})=0=\sin \pi \end{gathered}[/tex]The equations are:
[tex]\begin{gathered} x+\frac{\pi}{6}=0 \\ x+\frac{\pi}{6}=\pi \end{gathered}[/tex]We can substract π/6 from both sides of both equations:
[tex]\begin{gathered} x+\frac{\pi}{6}-\frac{\pi}{6}=0-\frac{\pi}{6} \\ x=-\frac{\pi}{6} \\ x+\frac{\pi}{6}-\frac{\pi}{6}=\pi-\frac{\pi}{6} \\ x=\frac{5\pi}{6} \end{gathered}[/tex]So we have a vertical asymptote at x=-π/6 and another one at x=5π/6. This means that we just need to graph f(x) between these two vertical lines. It is also important to note that f(x) reaches positive or negative values when the value of x approaches to -π/6 or 5π/6.
Now that we have the asymptotes let's find the x-intercept i.e. the point where f(x) meets with the x-axis. This happens when f(x)=0 which happens when the numerator is equal to 0. Then we get:
[tex]\cos (x+\frac{\pi}{6})=0[/tex]The cosine is equal to zero at π/2 so we have:
[tex]\begin{gathered} \cos (x+\frac{\pi}{6})=0=\cos \frac{\pi}{2} \\ x+\frac{\pi}{6}=\frac{\pi}{2} \end{gathered}[/tex]We can substract π/6 from both sides:
[tex]\begin{gathered} x+\frac{\pi}{6}-\frac{\pi}{6}=\frac{\pi}{2}-\frac{\pi}{6} \\ x=\frac{\pi}{3} \end{gathered}[/tex]So the x-intercept is located at x=π/3. So for now we have the x-intercept and two vertical asymptotes so at the moment we have the following:
The black dot is the x-intercept at (π/3,0) and the dashed lines are the asymptotes. Our function passes through the black dot and is limited by the asymptotes.
We still need to find if it reaches positive or negative infinite values when approaching to the asymptotes. As we saw the function is equal to zero at x=π/3. This means that between the first asymptote and x=π/3 the function is either entirely positive or entirely negative. The same happens with the interval between x=π/3 and the second asymptote. So we have two intervals where the function mantains its sign: (-π/6,π/3) and (π/3,5π/6). Let's evaluate f(x) in one value of each interval and see if it's positive or negative there. For example, x=0 is inside the first interval and x=2 is inside the second interval:
[tex]\begin{gathered} f(0)=1.73205>0 \\ f(2)=-1.4067<0 \end{gathered}[/tex]So f(x) is positive at (-π/6,π/3) which means that as x approaches to -π/6 from the right it reaches positive infinite values. We also have that f(x) is negative at (π/3,5π/6) so as x approaches 5π/6 from the left the function reaches negative infinite values.
Using this information and the fact that the graph must pass throug the x-intercept we can graph the function. It should look like this:
And that's the graph of f(x).
the answer is red show me how to get to the answer
The given expression is:
[tex]\frac{5\sqrt{4}}{\sqrt{3}}[/tex]The first step is to find the square root of 4 in the numerator, that is:
[tex]\sqrt{4}\text{ = 2}[/tex]Substitute this into the given expression:
[tex]\frac{5(2)}{\sqrt{3}}[/tex][tex]\frac{10}{\sqrt{3}}[/tex]The next step is to rationalize, that is, multiply the numerator and the denominator by √3
[tex]\frac{10}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}[/tex][tex]\frac{10\sqrt{3}}{\sqrt{9}}[/tex]Since √9 = 3[tex]\frac{10\sqrt{3}}{3}[/tex]How would I write an equation in point- slope form with inequalities, slope-intercept form with inequalities and standard form with inequalities with these three sets of points(9,7) (8,5)(2,9) (2,7)(3,5) (5,4)
a. The point-slope equation is:
[tex]y-y_1=m(x-x_1)[/tex]Where m is the slope and (x1,y1) are the coordinates of one point in the line. Also, you need to write the equation with inequalities, then you need to replace the = sign, for a <, > or <=, >= sign.
Let's start by finding the slope of the first set of points (9,7) (8,5).
The formula for the slope is:
[tex]m=\frac{y_2-y_1}{x_2-x_1_{}}[/tex]By replacing the values you obtain:
[tex]m=\frac{5-7_{}}{8_{}-9}=\frac{-2}{-1}=2[/tex]The slope is 2.
Now, replace this value into the slope-form equation and the values of the first point (9,7):
[tex]y-7_{}>2(x-9)[/tex]I choose the sign > (greater than), but you can choose anyone, the difference will be for the solution of the inequality. When you solve the inequality you will find that the x-values have to be greater than the solution you found, or less than... etc, it will depend on the sign you have in the inequality.
b. The slope-intercept equation is:
[tex]y=mx+b[/tex]Where m is the slope and b the y-intercept.
Let's use the second set of points (2,9) and (2,7)
Start by calculating the slope:
[tex]m=\frac{7-9}{2-2}=\frac{-2}{0}=\text{ undefined}[/tex]As there's no difference in the x-coordinates, the line is a vertical line at x=2.
Also, there's no y-intercept as the line never crosses the y-axis.
I will use the first set again, so you can understand the slope-intercept form.
From part a) you know that the slope is 2, let's replace it in the equation and use the first pair of coordinates to find b:
[tex]\begin{gathered} 7=2\times9+b \\ 7=18+b \\ 7-18=b \\ b=-11 \end{gathered}[/tex]Thus, the slope-intercept with inequality will be:
[tex]y<2x-11[/tex]c. The standard form equation of a line is:
[tex]ax+by=c[/tex]Let's use the third set of points (3,5) (5,4).
Start by finding the slope:
[tex]m=\frac{4-5}{5-3}=\frac{-1}{2}=-0.5[/tex]Now, you can start with the point-slope form and then convert it into the standard form:
[tex]\begin{gathered} y-5\ge-0.5(x-3) \\ Apply\text{ the distributive property} \\ y-5\ge-0.5x+1.5 \\ y\ge-0.5x+1.5+5 \\ y\ge-0.5x+6.5 \\ 0.5x+y\ge6.5 \end{gathered}[/tex]Where a=0.5, b=1 and c=6.5
9) Find the slope of the line that passes through these two points. (0.3) and (4, -2)
To find the slope of the line that passes through points (0, 3) and (4, -2), we can use the formula for the slope of a line:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]We have then:
(0, 3) ---> x1 = 0, y1 = 3
(4, -2) --->x2 = 4, y2 = -2
Therefore:
[tex]m=\frac{-2-3}{4-0}\Rightarrow m=\frac{-5}{4}\Rightarrow m=-\frac{5}{4}[/tex]Then, the slope of the line that passes through points (0, 3) and (4, -2) is m = -5/4.
Sophia spent $40 on supplies to make 20 bracelets. She plans to sell them at a craft show for $5 each. Let y represent the amount of her profit. Is it discrete or continuous? And what are the domain and range?
we have that
y ------> the amount of her profit
x -----ghe number of bracelets
REmember that
Profit is equal to sell minus cost
so
y=5x-40
the domain is the interval (0,1,2,3,4,5) ------> is a discrete
the range is equal to
For x=
difference between managerial and financial accounting?
The main difference between managerial accounting and financial accounting is who the statement is being prepared for.
What differentiates managerial and financial accounting?There are several things that differentiate managerial accounting from financial accounting but the main one is the target of the financial statements.
Financial accounting is used for financial statements for those outside the company such as investors, customers, and the government. As such, it has to abide by certain standards.
Managerial accounting on the other hand, is for managers and decision makers in the company. It is less restrictive as the main goal is to provide information for managers and not those outside.
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Your friend says, "If a quadrilateral has a pair of opposite sides that are congruent 6 points and a pair of opposite sides that are parallel, then it is a parallelogram." What is your friend's error? Explain.
to be a parallelogram we need to have the 2 pairs of sides to be parallel so the correct option is C
Question 23A company's logo was designed using circles of 3 different sizes. The diameters of two of the circles are shown6 cm12 cmWhich measurement is closest to the area of the largest circle in square centimeters?D2021 Illuminate Education Inc.
SOLUTION
A company's logo was designed using circles of 3 different sizes. The diameters of two of the circles are shown:
6 cm
12 cm
Which measurement is closest to the area of the largest circle in square centimeters?
The measurement is closest to the area of the largest circle in square centimeters is
12 cm since it has a radius of 6 cm with 36 pi square centimetres; unlike the diameter
of 6 cm which has 3 cm radius and 9 pi square centimetres.
The correct answer is 12 cm.
a farmer has 150 yards of fencing to place around a rectangular garden. the fence will have an opening that is 1/3 of the gardens length(see picture). write a function a(x) that describes the area of the garden.Find the dimensions of the garden if it has the maximum area, and find the maximum area.
By forming equations, we know that the garden is 37.5 yards long, and 37.5 yards wide, and it has an opening that is 12.5 yards wide.
What are equations?A mathematical equation is a formula that uses the equals sign to express the equality of two expressions. A mathematical statement that has an "equal to" symbol between two expressions with equal values is called an equation. As in 3x + 5 = 15, for instance. Equations come in a variety of forms, including linear, quadratic, cubic, and others.So, the dimensions and area of the garden:
Let x and y stand for length and width, respectively.There are 150 yards of fencing available, so:
2(x + y) = 150x + y = 75y = 75 - x ...(1)The garden's area (A) is given as follows:
A = xyA = x(75 - x)A = 75x - x²At A' = 0, the area is largest.
A' = 75 - 2x75 - 2x = 0x = 37.5 yardsy = 75 - x y = 75 - 37.5 = 37.5Garden opening: 1/3 × 37.5 = 12.5 yards
Therefore, by forming equations, we know that the garden is 37.5 yards long, and 37.5 yards wide, and it has an opening that is 12.5 yards wide.
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The correct question is given below:
A farmer has 150 yards of fencing to place around a rectangular garden. The fence will have an opening that is 1/3 of the garden's length. Write a function A(x) that describes the area of the garden, where x is the length of the garden. Find the dimensions if that has a maximum area, and find the maximum area
Answer:
The length is 45 yards.
The width is 37.5 yards.
The area is 1,687.5 yards
Step-by-step explanation:
I go to RSM too lolol
The other answer posted here was incorrect. Since the length and width DID have a relatively equal factor without the -12.5 yard opening taken into account, you'd usually get 37.5 yards.
BUT, if we use the width and subtract it from the total (which we'd get 75 yards left), we can see that in the total length, a sixth (it is 1/6th since we are taking both sides) is taken from that. 75 is easily dividable by 5, so we can take 15, and multiply it by 3. We'd then get 45 yards in total for each side (minus the 12.5 yard opening).
All you need to do now is multiply the length and width to get 1,687.5 yards.
Now get a 100 on that RSM assignment and get the bragging rights for your class. You can thank me later. Your homework is more important.
When a stone is dropped in a pond ripples are formed and travel in concentric circles away from where the stone was dropped. The relationship between the time and area if the circles was analyzed and is shown in the computer output.
What is the equation of the least/squares regression line?
When a stone is dropped in a body of water, ripples are created and move outward in concentric rings. The least-squares regression line has the equation Area = 0.010 + 3.141 (Time2).
The least squares regression equation is the equation y=1x+0 that specifies the least squares regression line.
the least squares regression line's y=1x+0 equation.
What is relationship between time and distance?Time (t = d/v) or, alternatively, time = speed/distance (v = d/t).
Concentric circles are those that share a common center but have differing radii. It is described as two or more circles with the same center, in other words. An annulus is the space between two concentric circles with dissimilar radii.
Concentric circles are two or more circles with the same center but various radii. Congruent circles are any two or more circles that have the same radius but different centers.
Concentric circles are those that share a common center but have differing radii. It is described as two or more circles with the same center, in other words.
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Translate the sentence into an inequality, the product of c and 9 is greater than 16.
In order to write an inequality we can read the original statement in small parts.
In this case, the statement is:
"the product of c and 9 is greater than 16"
We have that "the product" is a multiplication
Then, "the product of c and 9" is the multiplication between c and 9:
9 · c
And, "the product of c and 9 is greater than 16" means that 9 · c is greater than 16:
9 · c > 16
Answer: 9 · c > 16clarissa's division test was a 60% the first six weeks and a 72% the second six weeks. find the percent change.
The required change in the percentage is 20%.
Given that,
clarissa's division test was 60% in the first six weeks and 72% in the second six weeks. To determine the percent change.
The percentage is the ratio of the composition of matter to the overall composition of matter multiplied by 100.
Here,
According to the question,
The change in the percentage = (72 - 60) / 60 × 100
The change in the percentage = 20%
Thus, the required change in the percentage is 20%.
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I need help on 3 it says find the value of x round each answer to the nearest tenth
In problem 3, we have a right triangle with:
• cathetus ,a = 7,,
,• cathetus ,b = x,,
,• and hypotenuse ,h = 9,.
Pigatoras Theorem states that:
[tex]h^2=a^2+b^2.[/tex]Where a and b are cathetus and h the hypotenuse.
Replacing the data of the problem in the equation above, we have:
[tex]9^2=7^2+x^2.[/tex]Solving for x the last equation, we get:
[tex]\begin{gathered} 81=49+x^2, \\ x^2=81-49, \\ x^2=32, \\ x=\sqrt[]{32}\cong5.7. \end{gathered}[/tex]Answer
The value of x to the nearest tenth is 5.7.
Find the slope of the two points: (-3,-2) & (5, -8)
ter Numerical value ONLY. NO Decimals
*
[tex](\stackrel{x_1}{-3}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{-8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-8}-\stackrel{y1}{(-2)}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{(-3)}}} \implies \cfrac{-8 +2}{5 +3} \implies \cfrac{ -6 }{ 8 } \implies - \cfrac{ 3 }{ 4 }[/tex]
one box of strawberries cost $5 what equation can be used to calculate the most number of boxes a person can buy with $30
Answer
The maximum number of boxes that one can buy with 30 dollars is 6 boxes of strawberries.
Explanation
One box of strawberries cost 5 dollars.
If one buys x boxes of strawberries, the cost would be (5x) dollars.
So, if one has 30 dollars, we want to find the maximum number of boxes that the person can buy, that is, the maximum value of x
5x ≤ 30
Divide both sides by 5
(5x/5) ≤ (30/5)
x ≤ 6
The number of boxes that one can buy is less than or equal to 6.
Hence, the maximum number of boxes that one can buy with 30 dollars is 6 boxes of strawberries.
Hope this Helps!!!
Which of the following equations could you solve by first adding six and then dividing by negative three?
3x - 6 = -9
6 - 3x = -9
-3x - 6 = -9
x/-3 + 6 = -9
Answer:
-3x-6=-9
Step-by-step explanation:
[tex]-3x-6=-9[/tex]
[tex]+6[/tex] [tex]+6[/tex]
[tex]-3x = -3[/tex]
[tex]-3x/-3 = -3/-3[/tex]
[tex]x=1[/tex]
I already wrote the answer I just need you to work it out for me please and thank you
Answer:
[tex]A=470\frac{1}{4}ft^2[/tex]Detailed Explanation: The area of the figure provided is the sum of two areas, a rectangle, and a triangle:
The total area is calculated next, and the necessary steps are shown as follows
[tex]\begin{gathered} A=A_1+A_2 \\ A_1=\frac{1}{2}(b\cdot h)=\frac{1}{2}\cdot\lbrack(25ft-22.5ft)\times19.8ft\rbrack \\ A_1=\frac{1}{2}\cdot\lbrack2.5ft\times19.8ft\rbrack=\frac{49.5ft^2}{2}=24.75ft^2 \\ A_1=24.75ft^2 \\ A_2=w\cdot h=22.5ft\cdot19.8ft=445.5ft^2 \\ A_2=445.5ft^2 \\ \therefore\Rightarrow \\ A=A_1+A_2=24.75ft^2+445.5ft^2 \\ A=470.25ft^2 \\ A=470\frac{1}{4}ft^2 \end{gathered}[/tex]Which sequence of transformations will change figure PQRS to figure P'Q'R'S'?
Explanation:
A counterclockwise rotation about the origin by 90 degrees rule is:
[tex](x,y)\rightarrow(-y,x)[/tex]The reflection about the x-axis is:
[tex](x,y)\rightarrow(x,-y)[/tex]If we take for example point P (-3, -2) we can see it ends at P'(2,3). The counterclockwise rotation about the origin by 90º gives:
[tex](-3,-2)\rightarrow(2,-3)[/tex]And now with a reflection about the x-axis:
[tex](2,-3)\rightarrow(2,3)[/tex]Which is point P'
Answer:
Counterclockwise rotation about the origin by 90 degrees followed by reflection about the x-axis
exponent hw.simplify
Answer:
4
Explanation:
Given the below;
[tex]4m^0[/tex]To simplify the above, we have to note that any number or variable raised to the power of 0 is 1.
So, we'll have;
[tex]4m^0=4\times1=4[/tex]Hello this is a multi step question and I am struggling to help my son with this. It is 1 of 3 so hoping to get guidance with this first one to be able to know how to apply it to the others in his activities. Thank you as I know this is multiple steps and time consuming. The help is greatly appreciated as a parent.
In the first part of this problem, we must compute some statistic variables of two distributions:
0. the mean value,
,1. the median,
,2. the standard deviation.
,3. the interquartile range.
1. The mean of a data set is the sum of all the data divided by the count n:
[tex]\mu=\frac{x_1+x_2+\cdots+x_n}{n}\text{.}[/tex]2. The median is the data value separating the upper half of a data set from the lower half, it is computed following these steps:
• arrange data values from lowest to the highest value,
,• the median is the data value in the middle of the set
,• if there are 2 data values in the middle the median is the mean of those 2 values.
3. The standard deviation for a sample data set is given by the following formula:
[tex]\sigma=\sqrt[]{\frac{(x_1-\mu)^2+(_{}x_2-\mu)^2+\cdots+(x_n-\mu)^2}{n-1}_{}}\text{.}[/tex]4. The interquartile range (IQR) is given by:
[tex]\text{IQR}=Q_3-Q_1\text{.}[/tex]Where Q_1 and Q_3 are the first and third quartiles. The lowest quartile (Q1) covers the smallest quarter of values in your dataset.
--------------
Using the definitions above, we compute the mean, the median and the standard deviation for the samples taken by Manuel and Gretchen.
Manuel's sample
• Sample = {3, 6, 8, 11, 12, 8, 6, 3, 10, 5, 14, 9, 7, 10, 8}
,• Count = 15
1. Mean
Using the formula above, we get:
[tex]\mu=\frac{120}{5}=8.[/tex]2. Median
We order the data set:
[tex]3,3,5,6,6,7,8,(8),8,9,10,10,11,12,14.[/tex]From the ordered data set, we see that the central number 8 divides the data set into two equal parts.
So the median of this sample is:
[tex]\bar{x}=8.[/tex]3. Standard deviation
Using the formula above, we get:
[tex]\sigma=\sqrt[]{\frac{138}{15-1}}\cong3.14.[/tex]4. Interquartile range
Dividing the data sample into quartiles, we have:
[tex]3,3,5,6|6,7,8|8|8,9,10|10,11,12,14.[/tex]We have:
• Q_1 = 6,
,• Q_3 = 10.
So the interquartile range is:
[tex]\text{IQR }=Q_3-Q_1=10-6=4.[/tex]Gretchen's sample
• Sample = {22, 4, 7, 8, 12, 15, 10, 7, 9, 6, 13, 3, 8, 10, 10}
,• Count = 15
1. Mean
[tex]\mu=\frac{144}{15}=9.6.[/tex]2. Median
We order the data set:
[tex]3,4,6,7,7,8,8,(9),10,10,10,12,13,15,22.[/tex]From the ordered data set, we see that the central number 8 divides the data set into two equal parts.
So the median of this sample is:
[tex]\bar{x}=9.[/tex]3. Standard deviation
[tex]\sigma=\sqrt[]{\frac{307.6}{15-1}}\cong4.69.[/tex]4. Interquartile range
Dividing the data sample into quartiles, we have:
[tex]3,4,6,7|7,8,8|9|10,10,10|12,13,15,22.[/tex]We have:
• Q_1 = 7,
,• Q_3 = 12.
So the interquartile range is:
[tex]\text{IQR }=Q_3-Q_1=12-7=5.[/tex]Answers
Manuel's sample
0. Mean = 8
,1. Median = 8
,2. Standard deviation ≅ 3.14
,3. Interquartile range = 4
Gretchen's sample
0. Mean = 9.6
,1. Median = 9
,2. Standard deviation ≅ 4.69
,3. Interquartile range = 5
To get the variable r alone on one side of the equation below, Amy multiplied both sides of the equation by 4. is she correct? Explain why or why not. Solve the equation. 4r = 124
Given the equation
4r=124
You have to clear the value of r, this is, that r ends up alone in one side of the equation and the rest of the terms of the equation stay in the other side.
As you can see r is being multiplied by 4, to nullify this multiplication you have to "reverse the operation" that is, divide it by four.
And for the equality to continue, every operation made in one side of the equation has to be done in the other side, this means that if you divide 4r by 4, you have to divide 12
A parabola contains the following points.(-5,8),(2,-3),(3,1) which of the following systems of equations could be solved in order to find the equation that corresponds to this parabola?
Generic parabola equation:
y = a*x^2 + b*x + c
We have three points of the parabola:
(-5,8), (2,-3), (3,1)
For the point (-5, 8): x = -5, y = 8
8 = 25*a - 5*b + c
Point (2,-3): x = 2, y = -3
-3 = 4*a + 2*b + c
Point (3, 1): x = 3, y = 1
1 = 9*a + 3*b + c
Our system of equations:
8 = 25*a - 5*b + c
-3 = 4*a + 2*b + c
1 = 9*a + 3*b + c
The last option is the correct answer