By definition, you know that two equations are equivalent when they have the same solution. So, solving the first equation you have
[tex]\begin{gathered} \frac{1}{2}x-8=9 \\ \text{ Add 8 }to\text{ both sides of the equation} \\ \frac{1}{2}x-8+8=9+8 \\ \frac{1}{2}x=17 \\ \text{Multiply by 2 }on\text{ both sides of the equation} \\ 2\cdot\frac{1}{2}x=17\cdot2 \\ x=34 \end{gathered}[/tex]Now, solving the second equation you have
[tex]\begin{gathered} x-8=18 \\ \text{ Add 8 to both sides of the equation} \\ x-8+8=18+8 \\ x=26 \end{gathered}[/tex]Since the equations do not have the same solution then these equations are not equivalent.
Solve the inequality for A:-3 (9 – 4A) > 3 (2A – 11).
Given the inequality
-3 (9 – 4A) > 3 (2A – 11).
expand
-27 + 12A > 6A -33
Collect like terms
12A - 6A > -33 +27
6A > -6
Divide both sides by 6
A > -1
your freezer should be kept at -18 C. One day you woke up and noticed the door was left open and the temperature is now -4 degrees C. How many degrees warmer is the freezer now?
where are given that the reference temperature of a freezer is -18 C, if the temperature is -4 C. the difference in temperature will give us how many degrees warmer the freezer is:
[tex]\Delta T=-4-(-18)[/tex]To solve the operations we need first change the sing inside the parenthesis since it is preceded by a minus sing.
[tex]\Delta T=-4+18[/tex]Solving the operations:
[tex]\Delta T=14[/tex]Therefore, the freezer is 14C warmer.
Look at the graph below and use the vertical line test to determine whether or not the graph represents a function. The determine if it is a one-to-one function.positive cube root functionThis graph Answer represent a function.This graph Answer represent a one-to-one function.
By using the vertical line test with the given graph
Since a vertical line can intersect the graph in any position in only ONE point
Then the graph represents a function
To test if it is a one-one function, draw a horizontal line and check if it intersects the graph at any position in only ONE point or not
Since a horizontal line can intersect the given graph in only ONE point in any position, then
The graph represents a one-one function
A company charges (c) a flat fee of $4 for shipping plus $0.85 per pound (p). Which equation expresses this relatioship?
The cost of shipping is the flat rate plus the cost per pound times the number of pounds
c = flat rate + rate* pounds
c = 4 + 0.85*p
The write the second part in the opposite order
c = 0.85p +4
When we add the order doesn't matter
Answer B c = 0.85p + 4
Can I please get someone to help me with this?
First, notice that the line intersects the y-axis at the point (0,5), and that it also passes through the point (2,4). Then, we can use the equation for the slope given two points:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]in this case, we get the following:
[tex]\begin{gathered} (x_1,y_1)=(2,4)_{} \\ (x_2,y_2)=(0,5) \\ \Rightarrow m=\frac{5-4}{0-2}=-\frac{1}{2} \\ m=-\frac{1}{2} \end{gathered}[/tex]now that we have that the slope is m = -1/2 and that the y-intercept is 5 (since the intersection is the point (0,5)), then, the equation of the line in slope intercept form is:
[tex]y=-\frac{1}{2}x+5[/tex]the doll collector store has an inventory of 420 dolls a total of 70 dogs are made of porcelain and the remainder are made of plastic which of the following is the ratio of the plastic dolls to the total number of dolls in store inventory
let:
P = Number of plastic dolls = 420 - 70 = 350
N = total number of dolls in store inventory
[tex]\begin{gathered} P\colon N \\ 350\colon420=\frac{350}{420}=\frac{5}{6} \end{gathered}[/tex]how many 1/5s are in 20?
there are 100 1/5s in 20
If tan A = 21/20 and cos B = 28/53 and angles A and B are in Quadrant I, find the valueof tan(A - B).
Graph the function. f(x) = -3 sin x Use 3.14 for Use the sine tool to graph the function. The first point m value on the graph closest to the first point....I'll send pic of the problem
This is an example of how the graph should look like. What you can do is to find the sine of 2 angles, let's choose pi and 3pi/2. Find the function for these values
[tex]\begin{gathered} -3\cdot\sin \pi=-3\cdot0=0 \\ -3\cdot\sin (\frac{3\pi}{2})=-3\cdot-1=3 \end{gathered}[/tex]Plot these points and graph the function: (3.14,0) and (4.71,-3)
Our university consists of three colleges: business, engineering, and fine arts. There are 2,900 students in the business college, 1,500 students in the engineering college, and 1,000 students in the fine arts college. What percent of the total number of students are in the fine arts college. Round your answer to the nearest percent.
Given data:
The numbers of students in business college is B=2,900.
The numbers of students in engineering college is E=1,500.
The numbers of students in fine arts is A=1,000.
The percentage of total number of students in fine arts is,
[tex]\begin{gathered} P=\frac{A}{B+E+A}\times100 \\ =\frac{1,000}{2,900+1,500+1,000}\times100 \\ =18.52\text{ percent} \\ \approx18\text{ percent} \end{gathered}[/tex]Thus, the percentage of the students in fine arts is 18 %.
Statistics and probil
we know that
Minimum value=838
Maximum value=1443
Difference=1443-838=605
we have that
Lower Class Limit Upper-Class Limit
838 838+x
838+x 838+2x
838+2x 838+3x
838+3x 838+4x
838+4x 838+5x
838+5x 838+6x=1443
Find out the value of x
838+6x=1443
6x=1443-838
6x=605
x=100.83
therefore
the answer is
Lower Class Limit Upper-Class Limit
838 838+100.83=938.83
938.83 1039.66
1039.66 1140.49
1140.49 1241.32
1241.32 1342.15
1342.15 1443
Find out the frequency for each class
838-938.83 ----> (838,842) ---------> frequency=2 ok
938.83- 1039.66 -----> (945,1034,1025) --------> frequency=3 ok
1039.66-1140.49 -----> (1124,1136,1057,1130) ----> frequency=4 ok
1140.49-1241.32 -----> (1184) ----> frequency=1 ok
1241.32-1342.15 ----> (1247, 1249,1256) -----> frequency=3 ok
1342.15- 1443 -----> (1352,1439,1439,1368,1381,1342,1395) -----> frequency=7
can you please help me solve this? i can't solve this question.
To solve this question, we have to relate period (seconds to make a cycle) and its length.
We can relate them as:
[tex]T=2\pi\sqrt[]{\frac{L}{g}}[/tex]where g is the acceleration due to gravity and L the length of the pendulum.
If T1=2.00 and T2=1.99, we can relate them as:
[tex]\begin{gathered} \frac{T_2}{T_1}=\sqrt[]{\frac{L_2}{L_1}} \\ \frac{L_2}{L_1}=(\frac{T_2}{T_1})^2=(\frac{1.99}{2.00})^2=0.995^2=0.990025 \\ L_1=\frac{L_2}{0.990025}\approx1.01L_2 \end{gathered}[/tex]Then, we know that the length of should be 1% larger than it actually is.
As we do not know the actual length, we will use the first equation to calculate the actual length first and then the correct length for a period of 2 seconds.
[tex]\begin{gathered} T=2\pi\sqrt[]{\frac{L}{g}} \\ \frac{T}{2\pi}=\sqrt[]{\frac{L}{g}} \\ L=g(\frac{T}{2\pi})^2 \\ L=9.81\cdot(\frac{1.99}{2\cdot3.14})^2=9.81\cdot0.3167^2=9.81\cdot0.1=0.981\text{ m} \end{gathered}[/tex]NOTE: all the variables and constants are in meters and seconds.
As the correct length is 1% larger than 0.981 m, we can calculate the increase in length as:
[tex]\Delta L=0.01\cdot L_2=0.01\cdot0.981m=0.00981\text{ m}[/tex]Answer: 0.00981 m
Factor the following difference of squares. *Check for a GCF.
ANSWER
(x + 15)(x - 15)
EXPLANATION
The difference of squares is equivalent to the product of the sum and subtraction of the bases,
[tex]a^2-b^2=(a+b)(a-b)[/tex]So, to factor this difference of squares, we have to find the principal square roots of each term,
[tex]\begin{gathered} \sqrt[]{x^2}=x \\ \sqrt[]{225}=15 \end{gathered}[/tex]So this is,
[tex]x^2-225=x^2-15^2=(x+15)(x-15)[/tex]Hence, the factored form is (x + 15)(x - 15).
Draw the image of a triangle after a dilation with a scale factor of 2.
Let's begin by listing out the information given to us:
The vertices of the triangle is given as:
[tex](0,0),(0,5),(-4,2)[/tex]Dilation by a scale factor of 2 means the triangle will be enlarged, the coordinate of the vertices become:
[tex]\begin{gathered} (0,0)\rightarrow2(0,0)=(0,0) \\ (0,5)\rightarrow2(0,5)=(0,10) \\ (-4,2)\rightarrow2(-4,2)=(-8,4)_{} \end{gathered}[/tex]We will then graph this
Susan's television was damaged during her move and she decides to replace it. She finds the television she wants at the BigBox Store. She can buy the television on consignment for $982 with a 14% down payment. How much must Susan pay as down payment? Round your answer to the nearest cent. Do not include a dollar sign in your answer.
ANSWER
$137.48
EXPLANATION
We have to find the 14% of $982,
[tex]982\cdot\frac{14}{100}=982\cdot0.14=137.48[/tex]Hence, she must pay
$137.48
Consider the following random sample of data: 12, 24, 30, 15, 22, 5, 9, 3, 101, 20
SOLUTION
The given data in desending order is:
[tex]3,5,9,12,15,20,22,24,30,101[/tex]Recall that the median is the middle number of set of numbers arranged in ascending or descending order.
Notice that there are 10 data values Hence the area two middle value
The median is the average of the middle values:
[tex]\frac{15+20}{2}=17.5[/tex]Hence the median is 17.5
Recall that an outlier is a data point that differs significantly from other observations
Hence the outlier is 101.
Note that new data will become:
[tex]\begin{equation*} 3,5,9,12,15,20,22,24,30 \end{equation*}[/tex]Therefore the median is 15
How many ounces are in 10 1/2 pounds?1 pound = 16 ounces
hello
to solve this question, we simply need to equate this
[tex]\begin{gathered} 1\text{pound}=16\text{ounces} \\ 10\frac{1}{2}=x\text{ ounces} \\ \text{cross multiply both sides } \\ x\times1=10.5\times16 \\ x=168\text{ounces} \end{gathered}[/tex]from the calculation above, 10.5 pounds would be equal to 168 ounces
Tom goes fishing with Jason. Tom catches five trout and four catfish. Jason catches twice as many trout as Tom did. How many trout did Jason catch?
We know that Tom catches five trout and four catfish and we also know that Jason catches twice as many trout as Tom did.
Knowing that Jason catches twice as many trout as Tom did we must multiply the number of trouts that Tom caught (5 trouts) by 2
[tex]5\cdot2=10[/tex]So, Jason caught 10 trouts.
Use Cramer's Rule to solve the system. You may use the calculator for computations only - do not use any matrix functions. Show all work
Solution
Therefore the value of
[tex]\begin{gathered} x=-1 \\ y=\frac{5}{2}=2.5 \end{gathered}[/tex]A box is filled with 12 red cards 2 blue cards and 4 green cards a card is chosen at random from the box what is the probability that the card is not blue write your answer as a fraction in simplest form
There are 12+2+4=18 cards in total. The probability to get a not-blue card is
[tex]\begin{gathered} P(\text{not blue) = 1 - P(blue)} \\ P(\text{not blue) =}1-\frac{2}{18} \end{gathered}[/tex]which gives
[tex]\begin{gathered} P(\text{not blue})=1-\frac{1}{9} \\ P(\text{not blue})=\frac{9}{9}-\frac{1}{9} \\ P(\text{not blue})=\frac{9-1}{9} \\ P(\text{not blue})=\frac{8}{9} \end{gathered}[/tex]then, the probability that the card is not blue is 8/9.
The points U(−1,9), V(−1,5), and W(8,9) form a triangle.Plot the points then click the "Graph Triangle" button. Then find the perimeter of the triangle. Round your answer to the nearest tenth if necessary.
Remember that the coordinates of the points are written in the form (x,y), the first entry represents the distance over the horizontal axis and the second entry represents the distance over the vertical axis.
Plot the given points on the coordinate plane:
The length of the segment UV is 4, and the length of the segment UW is 9. Since the triangle VUW is a right triangle, use the Pythagorean Theorem to find the length of the segment VW:
[tex]\begin{gathered} VW=\sqrt{UV^2+UW^2} \\ \\ =\sqrt{4^2+9^2} \\ \\ =\sqrt{97} \\ \\ \approx9.849 \end{gathered}[/tex]Add the lengths of all the segments to find the perimeter of the triangle:
[tex]\begin{gathered} P=UV+VW+UW \\ \\ =4+9.849...+9 \\ \\ =22.849... \\ \\ \approx22.8 \end{gathered}[/tex]Therefore, to the nearest tenth, the perimeter of the triangle is 22.8.
TASK 8 Michael has made a scale drawing of his classroom. The scale for his drawing is 0.5 in.: 3 ft. a. The length of the classroom is 30 ft. The length of the room on the scale drawing is 6 in. Is this correct? Explain why or why not. b. One of the student tables is 6 ft long. How long should it be on the drawing? Explain how you got your answer. c. Write your own problem concerning Michael's drawing. Solve and explain your answers.
The scale drawing is
Inches : Feet
0.5 : 3
We need to find the length of the classroom on the drawing if it is 30 feet
Let us use the ratio above to find it
Inches : Feet
0.5 : 3
x : 30
by using cross multiplication
[tex]x\times3=0.5\times30[/tex]3x = 15
Divide both sides by 3 to find x
[tex]\frac{3x}{3}=\frac{15}{3}[/tex]x = 5
The length on the drawing must be 5 inches
a) 6 inches is incorrect because the length on the drawing must be 5 inches
b) The student is 6 ft long
let us use the ratio above to find his length on the drawing
Inches : Feet
0.5 : 3
y : 6
By using cross multiplication
[tex]\begin{gathered} y\times3=0.5\times6 \\ 3y=3 \end{gathered}[/tex]Divide both sides by 3 to find y
[tex]\begin{gathered} \frac{3y}{3}=\frac{3}{3} \\ y=1 \end{gathered}[/tex]b) his length on the drawing is 1 inch
for number c) choose any length by feet and use the ratio to find its length on the drawing
Your height is 8 feet
Let us find it in the drawing
Inches : Feet
0.5 : 3
h : 8
By using cross multiplication
[tex]\begin{gathered} h\times3=0.5\times8 \\ 3h=4 \end{gathered}[/tex]Divide both sides by 3 to find h
[tex]\begin{gathered} \frac{3h}{3}=\frac{4}{3} \\ h=\frac{4}{3} \end{gathered}[/tex]c) Your height on the drawing is 4/3 inches
20. Connie's pool has 50 cubic yards of water in it and is draining at a rate of 3 cubic yards per second. Paula's pool has 9 cubic yards of water currently in it and is filling at a rate of 4 cubic yards per second. After how many seconds will Connie's pool have less water than Paula's?
write the equation for the Connie's pool and Paula's pool
Connie's
[tex]y=50-3x[/tex]y=cubic yards of water remaining in the pool
x=time in seconds
Paula's
[tex]y=9+4x[/tex]y=cubic yards of water in the pool
x=time in seconds
write the inequality in order for connie's pool to have less water
[tex]\begin{gathered} 50-3x<9+4x \\ \end{gathered}[/tex]solve the inequality for x
[tex]\begin{gathered} 50-9<3x+4x \\ 41<7x \\ x>\frac{41}{7} \end{gathered}[/tex]After 41/7 seconds Connie's pool will have less water than Paula's.
What is the speed of a jet plane that flies 8100 km in 9 hours (in km/hr)
V = d/t
Speed = distance/time
V = 8100km/9hr = 900Km/hr
Answer:
V = 900 km/h
I took a screenshot I didn’t want to type it again
You have a 52 standard deck.
There are 4 suites on the deck: diamonds, hearts, spades, and clubs.
Each suite has 13 ranks: Ace, 2, 3, 4, 5, 6, 7, 8, 9, Jack, Queen, and King → This means that there are 4 cards with each rank on the deck.
The "9 of clubs is missing on your deck"
This means that:
1) Your deck has one less card, the total number of cards is 51.
2) Your deck has one less club, instead of 13 club cards, you have 12.
3) Your deck has one 9 less, which means that there are 3 nines on your deck.
a) You have to select one event, whose probability decreased due to the missing 9 of clubs.
For example, the event "you draw a card at random and it's a 9"
The expected probability of drawing a 9 of the deck can be determined as the number of nines divided by the number of cards on the deck:
[tex]\begin{gathered} P(9)=\frac{4}{52} \\ P(9)=\frac{1}{13} \\ P(9)=0.076\approx7.6\% \end{gathered}[/tex]But in reality, there is one 9 is missing from the deck, so you have 3 nines and 51 cards on the deck, its probability is:
[tex]\begin{gathered} P(9)=\frac{3}{51} \\ P(9)=\frac{1}{17} \\ P(9)=0.059\approx5.9\% \end{gathered}[/tex]The expected probability of drawing a card at random and the card being a 9 is 7.6%, but due to the missing card, the probability dropped to 5.9%.
This means that drawing a card at random and selecting a 9 is less likely than expected.
b) You have to select one event whose probability increased due to the missing card.
For example, the probability of drawing an Ace, knowing that the card is a club:
On a normal deck there are 13 clubs and "one Ace of clubs", the expected probability of drawing the ace, given that the card is a club can be determined as follows:
[tex]\begin{gathered} P(\text{Ace}|\text{Club)}=\frac{1}{13} \\ P(\text{Ace}|\text{Club)}=0.076\approx7.6\% \end{gathered}[/tex]But we are missing one club, which means that the total number of clubs is missing, so instead of having 13 clubs, we have twelve. The probability can be determined as follows:
[tex]\begin{gathered} P(\text{Ace}|\text{Club)}=\frac{1}{12} \\ P(\text{Ace}|\text{Club)}=0.083\approx8.3\% \end{gathered}[/tex]The expected probability of drawing the Ace, given that the card is a club, on a normal deck is 7.6%, but due to the missing 9 of clubs, this probability has increased to 8.3%.
So this event is more likely due to the missing card.
c) You have to select an event whose probability hasn't changed due to the missing card.
For example, the event "draw a card at random and is a Heart"
The expected probability of drawing a heart from the deck is equal to the quotient between the number of hearts and the total number of cards on the deck:
[tex]\begin{gathered} P(H)=\frac{13}{52} \\ P(H)=\frac{1}{4} \\ P(H)=0.25\approx25\% \end{gathered}[/tex]Your deck is missing one card, so there are 13 Hearts and a total of 51 cards, the probability can be determined as follows:
[tex]\begin{gathered} P(H)=\frac{13}{51} \\ P(H)\approx0.254\approx25.4\% \end{gathered}[/tex]The probability of drawing a heart is around 25% when the deck is complete or missing one card.
Find the reference angle for a rotation of 297º.
We have the following angle:
We know that the reference angle is the smallest angle that the terminal side of a given angle makes with the x-axis. In this case, we have the following reference angle:
To find the value of the reference angle, we substract 297 from 360:
[tex]x=360-297=63[/tex]therefore, the reference angle for a rotation of 297° is 63°
Write the correct system of inequalities, by first defining x and y, that correctly models the situation. Then write the inequalities and then graph the situation stated below. For your stock portfolio, you have at most $4000 that you want to use to buy stock in two companies. One is a construction company, the other is a biotech company. You want to have at least 2 times as much in the construction company as you do in the biotech company. System of inequalities:
SOLUTION
Let x be a construction company,
Let y be a biotech company
From the question, we have that :
[tex]\begin{gathered} x\text{ + y }\leq\text{ 4000}\ldots\ldots..\ldots\ldots\ldots\ldots..equ\text{ 1 } \\ x\text{ }\ge\text{ 2y}\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots equ\text{ 2} \end{gathered}[/tex]
To determine whether or not it is sensible to do a regression analysis, look atQuestion 20 options: the slope y-intercept scatter plot correlation
Explanation
Regression analysis is a set of statistical methods used to estimate relationships between a dependent variable and one or more independent variables.
Before one determines if one will do a regression analysis, we will have to check for the scatter plot
A scatter plot is a type of plot or mathematical diagram using Cartesian coordinates to display values for typically two variables for a set of data.
Thus, the answer is a scatter plot
Does the function f(x) or g(x) have a greater value at x=2? f(x)=4∙2^x
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
graph: g(x)
function: f(x) = 4 * 2 ^ x
Step 02:
greater value ==> x = 2:
graph: g(x)
x = 2 , y = 18
g(2) = 18
function: f(x) = 4 * 2 ^ x
[tex]f(2)\text{ = 4 }\cdot2^2=\text{ 4 }\cdot\text{ 4 = 16}[/tex]The answer is:
g(x) has a greater value at x = 2
Find the number of CDs that will produce maximum revenue.
Given data:
Price of CD is,
[tex]p(x)=90-\frac{x}{6}[/tex]The total revenue is,
[tex]R(x)=90x-\frac{x^2}{6}[/tex]First find the derivative of revenue function and then equate it to zero we have,
[tex]\begin{gathered} R^{\prime}(x)=0 \\ 90-\frac{2x}{6}=0 \end{gathered}[/tex][tex]\begin{gathered} \frac{x}{3}=90 \\ x=90\times3 \\ x=270 \end{gathered}[/tex]Now, to prove the maximize find the double derivative of revenue function
[tex]\begin{gathered} R^{\doubleprime}(x)<0 \\ \frac{-2}{6}=\frac{-1}{3}<0 \end{gathered}[/tex]Thus, 270 CD's will produce maximum revenue.
Answer: Option (c) that is 270.