In order to determine which data sets should be displayed on a stem display, you consider that the stem display is usefull in the cases in which you have data which can be grouped easily. For instance, for data set in which there are differents number with the same first digit(s).
According with the previous definition you can notice that the options E) and A) are the best options, because there are different number that can be grouped, for example, according to the first number.
For other options you have other situations, for option D) there is no way to group the data. For option C) there is only one number on each data, so, there wouldn't be leafs in the diagram, and the same applies to option B), the first number is the same in all data, then, there is no way to group.
solve. 45÷n=5 problem
66. WORKER EFFICIENCY An efficiency study of the morning shift at a certain factory indicates that an average worker who arrives on the job at 8:00 A.M. will have assembled f(x) = -x³ + 6x² + 15x television sets x hours later. How many sets will such a worker have assembled by 10:00 A.M.? [Hint: At 10:00 A.M., x = 2.] b. Ilow many sets will such a worker assemble between 9:00 and 10:00 A.M.?
Step-by-step explanation:
use differential calculus
Complete the equation for the circle with center (6,2) and radius 8.
The equation of the circle is :-
[tex]\begin{gathered} (x-6)^2+(y-2)^2=8^2 \\ (x-6)^2+(y-2)^2=64 \end{gathered}[/tex]Finding the Midpoint of a Line
Segment
To find the midpoint, M, of AB we can use
formula for finding point C. This works
because the M lies along AB and divides it in
a fixed ratio. So, if the midpoint of AB is point M, what must the ratio of a : b be? Since we know the ratio of a to b, we can substitute the values you wrote above back into the formula for finding a point along a line segment.
I need this done in 20 minutes please and thank you
An isosceles triangle in the one that has 2 equal sides, means that it also has two equal angles
this means that:
[tex]\begin{gathered} ifJK=KL \\ \text{then,}\measuredangle KJL=\measuredangle KLJ \\ \end{gathered}[/tex]using the properties of the triangle
[tex]\begin{gathered} \measuredangle KJL+\measuredangle KLJ+\measuredangle JKL=180 \\ 2\cdot\measuredangle KLJ+\measuredangle JKL=180 \\ 2\cdot34+\measuredangle JKL=180 \\ \measuredangle JKL=180-68 \\ \measuredangle JKL=112 \end{gathered}[/tex]The image point of A after a translation left 2 units and down 5 units is the pointB(-8, -11). Determine the coordinates of the pre-image point A.Submit Answer
Let the coordinates of the pre-image which is point A be (x,y)
The point after the translation left 2 units and down 5 units is B(-8, -11)
To get the x coordinate
we were tolt that the point was move 2 units left
So this implies x = -8+2 = -6
To get the y coordinates, we wre told that the point was moved down 5 units
This implies y = -11+ 5 = -6
Therefore, the coordinates of the pre-image point A is (-6, -6)
Given 5x + 2y=22 and that y=1, find x
In a linear equation both variables are dependent on each other
Then to find x
first reorder and put x as one term only
x = (22- 2y)/5
now replace y by its given value y= 1
then x= (22- 2•1)/5 = 20/5= 4
Word Problems: Define your variable, write and solve the equation. 9. You have $75 to spend at the grocery store. You get $23 in change. How much money do you spend? Define your variable: Equation: Answer:
ANSWER
Variable : how much money was spent (x)
Equation: x + 23 = 75
Answer: $52
EXPLANATION
Let the variable (amount of money spent) be x.
Initially, you had $75 and you spend some money (x) and you are left with $23 in change.
This means that the sum of the money you spent and the change you received is $75.
That is:
x + 23 = 75
That is the equation.
To find how much money you spent, we have to find the value of x by collecting like terms:
=> x = 75 - 23
x = $52
You spent $52.
5.Given the sample triangle below and the conditions a=3, c = _51, find:cot(A).
Depending on the angle we are analyzing on the right triangle, each side of it takes a different name. In this case, we are going to name them depending on the angle A. Then,
a: opposite side (to A)
b: adjacent side
c: hypotenuse
STEP 2: formula for cot(A)We know that the formula for cot(A) is:
[tex]\cot (A)=\frac{\text{adjacent}}{\text{opposite}}[/tex]Replacing it with a and b:
[tex]\begin{gathered} \cot (A)=\frac{\text{adjacent}}{\text{opposite}} \\ \downarrow \\ \cot (A)=\frac{b}{a} \end{gathered}[/tex]Since a = 3:
[tex]\cot (A)=\frac{b}{3}[/tex]STEP 3: finding bWe have an expression for cot(A) but we do not know its exact value yet. First we have to find the value of b to find it out.
We do this using the Pythagorean Theorem. Its formula is given by the equation:
[tex]c^2=a^2+b^2[/tex]Since
a = 3
and
c = √51
Then,
[tex]\begin{gathered} c^2=a^2+b^2 \\ \downarrow \\ \sqrt[]{51}^2=3^2+b^2 \\ 51=9+b^2 \end{gathered}[/tex]solving the equation for b:
[tex]\begin{gathered} 51=9+b^2 \\ \downarrow\text{ taking 9 to the left} \\ 51-9=b^2 \\ 42=b^2 \\ \downarrow square\text{ root of both sides} \\ \sqrt{42}=\sqrt{b^2}=b \\ \sqrt[]{42}=b \end{gathered}[/tex]Then,
b= √42
Therefore, the equation for cot(A) is:
[tex]\begin{gathered} \cot (A)=\frac{b}{3} \\ \downarrow \\ \cot (A)=\frac{\sqrt[]{42}}{3} \end{gathered}[/tex]Answer: DA child has an empty box that measures 4 inches by 6 inches by 3 inches. View the figure.What is the length of the longest pencil that will fit into the box, given that the length of the pencil must be a whole number of inches? Do not round until your final answer.
Solution
For this case we can do the following:
We can find the value of s on this way:
[tex]s=\sqrt[]{6^2+4^2}=\sqrt[]{52}=7.21[/tex]And solving for r we got:
[tex]r=\sqrt[]{6^2+3^2}=\sqrt[]{45}=6.71[/tex]Then the answer for this case would be:
[tex]\sqrt[]{52}=7.21[/tex]HELPPPP ME PLEASE I NEED TO KNOW THE ANDWER ASAP
Given the figure in the attached image.
The polygon has 4 sides, and all the four sides are equal.
but the four angles are not given to be equal.
Therefore, the polygon shown in the figure is a Rhombus
The points (1,7) and (7,5) fall on a particular line. What is its equation in point-slope form?
Use one of the specified points in your equation. Write your answer using integers, proper fractions, and improper fractions. Simplify all fractions.
Answer:
[tex]y-7=-\dfrac{1}{3}(x-1)[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{4.4cm}\underline{Slope Formula}\\\\Slope $(m)=\dfrac{y_2-y_1}{x_2-x_1}$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ \\are two points on the line.\\\end{minipage}}[/tex]
To find the equation of a line that passes through two given points, first find its slope by substituting the given points into the slope formula.
Define the points:
(x₁, y₁) = (1, 7)(x₂, y₂) = (7, 5)Substitute the points into the slope formula:
[tex]\implies m=\dfrac{5-7}{7-1}=\dfrac{-2}{6}=-\dfrac{1}{3}[/tex]
Therefore, the slope of the line is -¹/₃.
[tex]\boxed{\begin{minipage}{5.8 cm}\underline{Point-slope form of a linear equation}\\\\$y-y_1=m(x-x_1)$\\\\where:\\ \phantom{ww}$\bullet$ $m$ is the slope. \\ \phantom{ww}$\bullet$ $(x_1,y_1)$ is a point on the line.\\\end{minipage}}[/tex]
To find the equation in point-slope form, simply substitute the found slope and one of the given points into the point-slope formula:
[tex]\implies y-7=-\dfrac{1}{3}(x-1)[/tex]
Lena eats an apple every otherday. Suppose today is Monday,October 1. Lena eats an appletoday.When will Lena eat an appleon a Monday again?AnsLe
There are 6000 students at Mountain High School, and 1/4 of these students are seniors. If 2/3 of the seniors are in favor of the school forming a debate team and 1/5 of the remaining students (not seniors) are also in favor of forming a debate team how many students do not favor this idea?
hello
tp solve this question, let's get the data out
we have a total of 6000 students. 1/4 of the students are seniors, let's find the numbers of seniors in the school
the number of senior students are
[tex]\frac{1}{4}\times6000=1500[/tex]we have 1500 seniors in the school.
we can find the seniors in support of a debate team by multiplying 2/3 by 1500
[tex]\frac{2}{3}\times1500=1000[/tex]now we know that 1000 students are in support of the debate team. let's subtract the numbers of students in support of debate team from numbers of seniors in the schoolto giveus the numbers of seniors that are not in support of debate team.
[tex]1500-1000=500[/tex]500 seniors from the school are not in support of a debate team.
Also note that 1/5 of the remaining students are not in support of the debate team.
which would be
[tex]\begin{gathered} 6000-1500=4500 \\ \frac{1}{5}\times4500=900 \end{gathered}[/tex]now, we can add the numbers of seniors that are not in support of a debate team plus number of remaining students not in support of a debate team
[tex]500+900=1400[/tex]from the calculationabove, a total of 1400 students are not in support of the idea
Given parallelogram ABCD, diagonals AC and BD intersect at point E. AE=2x, BE=y+10, CE=x+2 and DE=4y−8. Find the length of AC.A. 8B. 6C. 2D. 4
Answer:
the length of the diagonal AC is;
[tex]8[/tex]Explanation:
Given the parallelogram ABCD, diagonals AC and BD intersect at point E.
[tex]\begin{gathered} AE=2x \\ CE=x+2 \\ BE=y+10 \\ DE=4y+8 \end{gathered}[/tex]Recall that the diagonals of a parallelogram bisect each other;
So;
[tex]AE=CE[/tex]substituting AE and CE;
[tex]\begin{gathered} 2x=x+2 \\ 2x-x=2 \\ x=2 \end{gathered}[/tex]To calculate the length of AC;
[tex]\begin{gathered} AC=2x+x+2=3x+2 \\ since\text{ x=2} \\ AC=3x+2=3(2)+2 \\ AC=6+2 \\ AC=8 \end{gathered}[/tex]Therefore, the length of the diagonal AC is;
[tex]8[/tex]I would like some help on how to solve this problem!(please and thank you)
Answer
GH is congruent to JH and FH is congruent to HI. ∠GHF should be congruent to ∠JHI by Vertical Angles Theorem. Since GH is congruent to JH, ∠GHF is congruent ∠JHI, and FH is congruent is congruent to HI, ΔGHF is congruent ΔJHI, by SAS. Then one can assume that FG is congruent to IJ by CPCTC.
a teacher bought 4 folders and 9 books for $33.75. on another day, she bought 3 folders and 12 books at the same prices for $34.50. how much did she pay for each folder and each book?
The teacher made two different purchases:
First purchase:
4 folders and 9 books for $33.75
Second purchase
3 folders and 12 books for $34.50
Let "f" represent the cost of each folder and "b" represent the cost of each book. You can express the total cost of each purchase as equations:
[tex]\begin{gathered} 1)4f+9b=33.75 \\ 2)3f+12b=34.50 \end{gathered}[/tex]Now we have established a system of equations, to solve it, the first step is to write one of the equations in terms of one of the variables.
For example, I will write the first equation in terms if "f"
[tex]\begin{gathered} 4f+9b=33.75 \\ 4f=33.75-9b \\ \frac{4f}{4}=\frac{33.75-9b}{4} \\ f=\frac{135}{16}-\frac{9}{4}b \end{gathered}[/tex]The second step is to replace the expression obtained for "f" in the second equation:
[tex]\begin{gathered} 3f+12b=34.50 \\ 3(\frac{135}{16}-\frac{9}{4}b)+12b=34.50 \end{gathered}[/tex]Distribute the multiplication on the parentheses term
[tex]\begin{gathered} 3\cdot\frac{135}{16}-3\cdot\frac{9}{4}b+12b=34.50 \\ \frac{405}{16}-\frac{27}{4}b+12b=34.50 \\ \frac{405}{16}+\frac{21}{4}b=34.50 \end{gathered}[/tex]Pass the number to the right side of the equal sign by applying the opposite operation to both sides of it
[tex]\begin{gathered} \frac{405}{16}-\frac{405}{16}+\frac{21}{4}b=34.50-\frac{405}{16} \\ \frac{21}{4}b=\frac{147}{16} \end{gathered}[/tex]Now divide b by 21/4 to cancel the multiplication and to keep the equality valid, you have to divide both sides of the expression, so divide 147/16 by 21/4 too, or multiply them by its reciprocal fraction, 4/21, is the same.
[tex]\begin{gathered} (\frac{21}{4}\cdot\frac{4}{21})b=(\frac{4}{21}\cdot\frac{147}{16}) \\ b=\frac{7}{4}\approx1.75 \end{gathered}[/tex]Each book costs $1.75
Now that we have determined how much does each book cost, we can determine the cost of each folder by replacing the value of "b" in the expression obtained for "f"
[tex]\begin{gathered} f=\frac{135}{16}-\frac{9}{4}b \\ f=\frac{135}{16}-\frac{9}{4}\cdot\frac{7}{4} \\ f=\frac{9}{2}\approx4.5 \end{gathered}[/tex]Each folder costs $4.50
Find the set A n Φ.U = {1, 2, 3, 4, 5, 6, 7, 8, 9)A = 2, 3, 8, 9)Selectthe correct choice below and, if necessary, fill in the
Answer
Option B is the correct answer.
A n Φ = {}
A n Φ is the empty set.
Explanation
We are told to find the intersection between set A and the empty set Φ.
The intersection of two sets refers to the elements that belong to the two sets, that is, the elements that they both have in common.
Set A = (2, 3, 8, 9)
Set Φ = {}
What the elements of set A and set Φ (an empty set) have in common is nothing.
Hence, the intersection of set A and set Φ is an empty set.
A n Φ = {}
Hope this Helps!!!
The endpoints of the line are (0, 5) and (6, 4). Find the slope of the line.
Solution:
Given the endpoints of the line;
[tex](0,5),(6,4)[/tex]The slope, m of the line is;
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ \\ \text{ Where }x_1=0,y_1=5,x_2=6,y_2=4 \end{gathered}[/tex]Thus;
[tex]\begin{gathered} m=\frac{4-5}{6-6} \\ \\ m=-\frac{1}{6} \end{gathered}[/tex]CORRECT ANSWER:
[tex]-\frac{1}{6}[/tex]An electronics store makes a profit of $59 for everystandard DVD player sold and $69 for every portableDVD player sold. The manager's target is to make atleast $345 a day on sales from standard and portableDVD players. Write an inequality that represents thenumbers of both kinds of DVD players that can besold to reach or beat the sales target. Let s representthe number of standard DVD players sold and prepresent the number of portable DVD players sold.Then graph the inequality.
The profit on one standard DVD player is $59 and on one portable DVD player is $69.
If there are s number of standard DVD player then total profit on standard DVD players is $59s. Simillarly total profit on portable DVD players is $69p.
The total profit on DVD player shoul be at least $345, which means total profit on DVD players is $345 or more than $345.
The linear inequalty for total profit is,
[tex]59s+69p\ge345[/tex]The graph of the linear inequality is,
In graph, lines pointing away the origin represent the region for the equation.
A rectangular athletic field is twice as long as it is wide if the perimeter of the athletic field is 360 yards what are its dimensions. The width isThe length is
Step 1. We will start by making a diagram of the situation.
Since the length of the rectangle is twice the width, if we call the width x, then the length will be 2x as shown in the diagram:
Step 2. One thing that we know about the rectangle is its perimeter:
[tex]\text{Perimeter}\longrightarrow360\text{yd}[/tex]This perimeter has to be the result of the sum of all of the sides of the rectangle:
[tex]x+x+2x+2x=360[/tex]Step 3. Solve the previous equation for x.
In order to solve for x, the first step is to combine the like terms on the left-hand side:
[tex]6x=360[/tex]The second step to solve for x is to divide both sides of the equation by 6:
[tex]\frac{6x}{6}=\frac{360}{6}[/tex]Simplifying:
[tex]x=60[/tex]Step 4. Remember from the diagram from step 1, that x was the width of the rectangle:
[tex]\text{width}\longrightarrow x\longrightarrow60yd[/tex]and the length was 2x, so we multiply the result for the with by 2:
[tex]\text{length}\longrightarrow2x=2(60)=120\longrightarrow120yd[/tex]And these are the values for the width and the length.
Answer:
The width is 60yd
The length is 120yd
if a driver drive at aconstant rate of 38 miles per hour how long would it take the driver to drive 209 mile
In order to calculate how long would it take to drive 209 miles, we just need to divide this total amount of miles by the speed of the driver.
So we have:
[tex]\text{time}=\frac{209}{38}=5.5[/tex]So it would take 5.5 hours (5 hours and 30 minutes).
Line segments, AB,BC,CD,DA create the quadrilateral graphed on the coordinate grid above. The equations for two of the four line segments are given below. Use the equations of the line segments to answer the questions that follow. AB: y = -x + 1 BC: y = -3x + 11
The equations of the line segments are,
[tex]\begin{gathered} AB\colon y=\frac{1}{3}x+1 \\ BC\colon y=-3x+11 \end{gathered}[/tex]Calculate the equations of CD and AD.
The equation of line Cd is,
[tex]\begin{gathered} (y-(-3))=\frac{-1+3}{4+2}(x+2) \\ y+3=\frac{1}{3}(x+2) \\ 3y=x-7 \end{gathered}[/tex]The equation of the line AD is,
[tex]\begin{gathered} y-0=\frac{-3-0}{-2+3}(x+3) \\ y=-3x-9 \end{gathered}[/tex]1)If two lines are parallel slope will be equal and perpendicular product of slope will be -1.
From the equation, the slope of AB is 1/3
From the equation, the slope of Cd is 1/3.
So, they are parallel.
2)The slope of AB is 1/3.
The slope of BC is -3.
The product of two slopes is -1. Therefore, AB is perpendicular to BC.
3) The slope of AB is 1/3 and slope of AD is -3. Since, the product is -1, they are perpendicular.
Another pair of line segments that are perpendicular to each other is AB and AD.
a gate that is 5 ft tall casts a shadow 9 ft long the house behind the gate cast a shadow of 54 ft how about how many feet tall is the house
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data
gate:
hg = 5 ft
shadow = 9ft
house
hh = ?
shadow = 54 ft
Step 02:
We must apply the theorem of thales.
[tex]\frac{hg}{hh}=\frac{shadow\text{ gate}}{\text{shadow house}}[/tex][tex]\frac{5ft}{hh}=\frac{9ft}{54\text{ ft}}[/tex]hh * 9 ft = 5 ft * 54 ft
hh = (5 ft * 54 ft ) / 9 ft
hh = 270 ft ² / 9 ft = 30 ft
The answer is:
The house is 30ft tall.
Find the surface area of a right cone with diameter 30 in. and slant height 8 in.Your answerEXTRA CREDIT: Find the surface area of the figure below. Round to the nearesttenth, if necessary.10 in?
Answer:
Surface area = 1084 in²
Step-by-step explanation:
To find the surface area of a right cone, we can use the following formula:
[tex]\boxed{{Area = \pi r^2 + \pi rl}}[/tex],
where:
• r = radius
• l = slant height.
In the question, we are told that the diameter of the cone is 30 in. Therefore its radius is (30 ÷ 2 = ) 15 in. We are also told that its height is 8 in.
Using this information and the formula above, we can calculate the surface area of the cone:
Surface area = [tex]\pi \times (15)^2 + \pi \times 15 \times 8[/tex]
= [tex]345 \pi[/tex]
[tex]\approx[/tex] 1084 in²
Philip departed from town A with coordinates (1,6) towards town B with coordinates (7 ,6). At the same time Bruce headed from town B to town A. What are the coordinates of Point C where they will meet if the ration of Phillip's to Bruce's rates is 7:5 respectively ?
if there was no ratio they were in the middle (4,6)
but in this case we must multiply by the ratio
so
[tex]4\times\frac{7}{5}=\frac{28}{5}\approx5.6[/tex]so the C point is
[tex](5.6,6)[/tex]
Ali borrowed Php22,000 for 3months at the discount rate of 5 ¼ % from a bank. Find the (a) bank’s discount and (b) proceeds.
If an M amount is borrowed for a time t at a discount rate of r per year, then the discount D is calculated as
[tex]\begin{gathered} D=M\cdot r\cdot t \\ \\ \text{where} \\ r\text{ is expressed in decimals} \end{gathered}[/tex][tex]\begin{gathered} \text{Given} \\ M=22000 \\ r=5\frac{1}{4}\%\rightarrow5.25\%\rightarrow0.0525 \\ t=\mleft(\frac{3}{12}\mright)\text{or }0.25\text{ (3 months out of 1 year or 12 months} \end{gathered}[/tex]Substitute the following values to get the bank's discount.
[tex]\begin{gathered} D=Mrt \\ D=(22000)(0.0525)(\frac{3}{12}) \\ D=288.75 \end{gathered}[/tex]Therefore, the bank's discount is Php 288.75.
To calculate for proceeds, subtract the amount borrowed by the bank's discount.
[tex]\begin{gathered} P=M-D \\ P=22000-288.75 \\ P=21711.25 \end{gathered}[/tex]The proceeds given to Ali is Php 21,711.25.
The seventh term of a geometric sequence is 1/4 The common ratio 1/2 is What is the first term of the sequence?
Answer:
16
Explanation:
The equation for the term number n on a geometric sequence can be calculated as:
[tex]a_n=a_{}\cdot r^{n-1}[/tex]Where r is the common ratio and a is the first term of the sequence.
So, if the seventh term of the sequence is 1/4 we can replace n by 7, r by 1/2, and aₙ by 1/4 to get:
[tex]\frac{1}{4}=a\cdot(\frac{1}{2})^{7-1}[/tex]Then, solving for a, we get:
[tex]\begin{gathered} \frac{1}{4}=a(\frac{1}{2})^6 \\ \frac{1}{4}=a(\frac{1}{64}) \\ \frac{1}{4}\cdot64=a\cdot\frac{1}{64}\cdot64 \\ 16=a \end{gathered}[/tex]So, the first term of the sequence is 16.
1. describe the end behavior. 2. determine whether it represents an odd degree or an even degree function.3. state the number of real zeroes
1. Quadratic curve
2. Odd degree function
3. TWO REAL ZEROS
I need help on my practice sheet. needs to be simplified
then
[tex]\begin{gathered} \frac{x+6}{3x}\times\frac{3(x-6)}{(x+6)(x-6)} \\ \frac{x+6}{3x}\times\frac{3}{x+6} \\ \frac{(x+6)\times3}{3x\times(x+6)} \\ \frac{3}{3x} \\ \frac{1}{x} \end{gathered}[/tex]answer: 1/x
