The selection of three dogs out of 31 dogs can be done in
[tex]31C3\text{ ways}[/tex]i.e.
[tex]\begin{gathered} =\frac{31!}{(31-3)!\times3!} \\ =\frac{31\times29\times28!}{28!\times3!} \\ =\frac{31\times29}{6} \\ =149.8 \\ \cong\text{ 150} \end{gathered}[/tex]Owners of a recreation area are filling a small pond with water. They are adding water at a rate of 29 L per minute. There are 400 L in the pond to start. Let W represent the total amount of water in the pond (in liters) and let T represent the total number of minutes that water has been added.Write an equation relating W to T. Then use this equation to find the total amount of water after 13 minutes.Equation : Total amount of water after 13 minutes : liters
In this problem, we have a linear equation of the form
W=mT+b ----> equation in slope-intercept form
where
m is the unit rate or slope of the linear equation
m=29 L/min ----> given
b is the initial value
b=400 L ----> given
substitute
W=29T+400 -------> equation relating W to T.For T=13 min
substitute
W=29(13)+400
W=777 L
the total amount of water after 13 minutes is 777 Lcan some one clarify this question, i think ik the answer but i need some elses opinion Find m
hello
to solve this question, we simply need to add two quadrants that make up mto get m[tex]\begin{gathered} m<\text{WYV}=60^0 \\ m<\text{VYU}=85^0 \end{gathered}[/tex][tex]\begin{gathered} m<\text{WYU}=<\text{WYV}+m<\text{VYU} \\ m<\text{WYU}=60^0+85^0 \\ m<\text{WYU}=145^0 \end{gathered}[/tex]from the calculations above, the value of m
A survey of 100 high school students provided thisfrequency table on how students get to school:Drive toTake theGradeWalkSchoolbusSophomore2253Junior13202Senior2555Find the probability that a randomly selected studenteither takes the bus or walks.[?P(Take the bus U Walk)
Let's call the event of a student taking the bus as event A, and the event of a student walking as event B. The theoretical probability is defined as the ratio of the number of favourable outcomes to the number of possible outcomes. We have a total of 100 students, where 50 of them take the bus and 10 of them walk. This gives to us the following informations:
[tex]\begin{gathered} P(A)=\frac{50}{100} \\ P(B)=\frac{10}{100} \end{gathered}[/tex]The additive property of probability tells us that:
[tex]P(A\:or\:B)=P(A)+P(B)-P(A\:and\:B)[/tex]Since our events are mutually exclusive(the student either walks or takes the bus), we have:
[tex]P(A\:and\:B)=0[/tex]Then, our probability is:
[tex]P(A\cup B)=\frac{50}{100}+\frac{10}{100}-0=\frac{60}{100}=\frac{3}{5}[/tex]The answer is:
[tex]P(Take\:the\:bus\cup Walk)=\frac{3}{5}[/tex]8.1 km to miles and feet
Given
[tex]8.1\operatorname{km}[/tex]It should be noted that
[tex]\begin{gathered} 1\operatorname{km}=0.621371miles \\ 1\text{mile}=5280\text{feet} \end{gathered}[/tex][tex]\begin{gathered} \text{convert 8.1km to miles} \\ 1\operatorname{km}=0.621371\text{miles} \\ 8.1\operatorname{km}=8.1\times0.621371 \\ 8.1\operatorname{km}=5.0331051\text{miles} \end{gathered}[/tex][tex]\begin{gathered} 8.1\operatorname{km}=5\text{miles}+0.0331051\text{miles} \\ \text{convert 0.0331051miles to fe}et \\ 1\text{miles}=5280ft \\ 0.0331051\text{miles}=0.0331051\times5280feet \\ 0.0331051\text{miles}=174.79feet \end{gathered}[/tex]Hence, 8.1km is 5 miles and 174.79 feet
Find the volume of the given solid.Round to the nearest 10th, If necessary. In cubic inches
ANSWER
33.5 cubic inches
EXPLANATION
This is a cone with radius r = 2 in and height h = 8 in. The volume of a cone is,
[tex]V=\frac{1}{3}\cdot\pi\cdot r^2\cdot h[/tex]Replace the known values and solve,
[tex]V=\frac{1}{3}\cdot\pi\cdot2^2in^2\cdot8in=\frac{32}{3}\pi\text{ }in^3\approx33.5\text{ }in^3[/tex]Hence, the volume of the cone is 33.5 in³, rounded to the nearest tenth.
A 5p coin weighs 4.2g. Approximately, how much will one million pounds worth of 5p pieces
weigh?
Answer:
It would weight 840,000g
Step-by-step explanation:
1,000,000 ÷ 5
= 200,000
= 200,000 × 4.2
= 840,000
if x=10 units, then what is the volume of the cube
Knowing that the solid is a cube, you can use the following formula for calculate its volume:
[tex]V=s^3[/tex]Where "s" is the length of any edge of the cube.
In this case, you can identify that:
[tex]s=x=10units[/tex]are figures A and B congruent? explain your reason
Because of damage, a computer company had 5 tablets returned out of the 80 that were sold. Suppose the number of damaged tablets sold continue at this rate. How many tablets should the company expect to have returned if it sells 400 of them?
we are told that there 5 damaged tablets out of 80 that are sold. Therefore, the rate of damaged tablets per sold tablets is:
[tex]\frac{5\text{ damaged}}{80\text{ sold}}[/tex]Multiplying this rate by the 400 sold tablets we get:
[tex]\frac{5\text{ damaged}}{80\text{ sold}}\times40\text{0 sold}[/tex]Solving we get:
[tex]\frac{5\text{ damaged}}{80\text{ sold}}\times40\text{0 sold}=25\text{ damaged}[/tex]Therefore, if the rate continues, the company can expect to return 25 tablets.
A building is 5 feet tall. the base of the ladder is 8 feet from the building. how tall must a ladder be to reach the top of the building? explain your reasoning.show your work. round to the nearest tenth if necessary.
The ladder must be 9.4 ft to reach the top of the building
Here, we want to get the length of the ladder that will reach the top of the building
Firstly, we need a diagrammatic representation
We have this as;
As we can see, we have a right triangle with the hypotenuse being the length of the ladder
We simply will make use of Pythagoras' theorem which states that the square of the hypotenuse is equal to the sum of the squares of the two other sides
Thus, we have;
[tex]\begin{gathered} x^2=5^2+8^2 \\ x^2=\text{ 25 + 64} \\ x^2\text{ = 89} \\ x=\text{ }\sqrt[]{89} \\ x\text{ = 9.4 ft} \end{gathered}[/tex]Can I have help with this problem? I don't really understand how to graph this
Step 1:
The graph of y = -2 is a horizontal line passing through -2.
Step 2
Jan and her brother mel go to different schools. Jan goes 6 kilometer east from home. Mel goes 8 kilometer north. How many kilometer apart are their schools.
Jan goes 6 km east from her home, and Mel goes 8 km north from the same home. So we need to find the distance between both schools, and we do such using the Pithagorean theorem, since we are in the presence of a right angle triangle for which we know the two legs, and need to find the measure of the hypotenuse.
I am going to represent the problem with a diagram below, so you see the right angle triangle I am talking about.
So the two legs are represented by the distances each student travels, and the segment in red is the distance between the schools which appears as the HYPOTENUSE of the right angle triangle.
Therefore, we use the Pythagoras theorem for the hypotenuse:
[tex]\begin{gathered} \text{Hypotenuse}=\sqrt[]{leg1^2+\text{leg}2^2} \\ \text{hypotenuse}=\sqrt[]{6^2+8^2} \\ \text{Hypotenuse}=\sqrt[]{36+64} \\ \text{Hypotenuse}=\sqrt[]{100}=10 \end{gathered}[/tex]Therefore, the distance between the schools is 10 km
Question #3 3) The digits of a 2-digit number differ by 5. If the digits are interchanged and the resulting number is added to the original number, we get 99. Find the original number.
ones number = x
Tens number = y
y>x
Number at the tens place y = (x+5)
original number = 10 (x+5)+x
Interchange digits:= 10x+(x+5)
original number + new number = 99
¨[10(x+5)+x]+ [10x+ (x+5)] =99
Solving for x:
(10x+50+x )+( 10x+x+5) = 99
Combine like terms
(11x+50) + (11x+5) = 99
11x+11x+50+5 =99
22x+55 =99
subtract 55 from both sides
22x +55-55= 99-55
22x = 44
Divide both sides by 22
22x/22= 44/22
x = 2
unit place: 2
tens place = x+5 = 2+5 = 7
original number = 72
1. Identify the vertex (locator point) of the above parabola2 po(1,2)(3,0)(3,0)(2,1)2. Identify the vertex from the quadratic function y=-5(x-6) 2+82 point
Answer:
(2,1)
Step-by-step explanation:
The vertex of a parabola is it's highest point(if it is concave down), or it's lowest point, if it's concave up.
In this question:
It's concave down, so the vertex is the highest point.
It happens when x = 2, at which y = 1.
So the vertex is the point (2,1)
Larry purchased a new combine that cost $260,500, minus a rebate of $5,500, a trade-in of $8,500, and a down payment of $7,000. He takes out a loan for the balance at 8% APR over 4 years. Find the annual payment. (Simplify your answer completely. Round your answer to the nearest cent.)
The annual payment for the loan balance is $72,310.03.
What is the periodic payment?The periodic payment is the amount that is paid per period (yearly, monthly, quarterly, or weekly) to repay a loan or a debt.
The periodic payment can be computed using an online finance calculator, making the following inputs.
N (# of periods) = 4 years
I/Y (Interest per year) = 8%
PV (Present Value) = $239,500 ($260,500 - $5,500 - $8,500 - $7,000)
FV (Future Value) = $0
Results:
PMT = $72,310.03
Sum of all periodic payments = $289,240.13
Total Interest = $49,740.13
Thus, the annual payment that Larry needs to make is $72,310.03.
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A box of a granola contains 16.8 ounces . It cost $5.19 . What is the cost , to the nearest cent , of the granola per ounce ? A . $0.12 B . $0.31 C . $3.24
The cost per unit ounce is obtained by computing the quotient:
[tex]c=\frac{C}{N}.[/tex]Where:
• c is the cost per unit ounce,
,• C is the cost,
,• N is the number of ounces that you get for C.
In this problem we have:
• C = $5.19,
,• N = 16.8 ounces.
Computing the quotient, we get:
[tex]c=\frac{5.19}{16.8}\cong0.31[/tex]dollars per ounce.
Answer: B. $0.31
P(x) =x and q(x) = x-1Given:minimum x and Maximum x: -9.4 and 9.4minimum y and maximum y: -6.2 and 6.2Using the rational function [y=P(x)/q(x)], draw a graph and answer the following: a) what are the zeroes?b) are there any asymptotes? c) what is the domain and range for this function?d) it it a continuous function?e) are there any values of y= f(x)/g(x) that are undefined? Explain
we have the following function
[tex]\frac{p(x)}{g(x)}=\frac{x}{x\text{ -1}}[/tex]where x is between -9.4 and 9.4 and y is between -6.2 and 6.2.
We will first draw the function
from the graph, we can see that the zeroes are all values of x for which the graph crosses the x -axis
In this case, we see that that the only zero is at x=0.
Now, we have that the asymptotes are lines to which the graph of the function get really close to. On one side, we can see that as x goes to infinity or minus infinity, the values of the function get really close to 1. So the graph has a horizontal asymptote at y=1. Also, we can see that as x gets really close to 1, the graph gets really close to the vertical line x=1. So the graph has a vertical asymptote at x=1.
Recall that the domain of a function is the set of values of x for which the function is defined. From our graph, we can see that graph is not defined when x=1. So the domain of the function is the set of real numbers except x=1. Now, recall that the range of the function is the set of y values of the graph. From the picture we can see that the graph has a y coordinate for every value of y except for y=1. So, this means that the range of the function is the set of real numbers except y=1.
From the graph, we can see that we cannot draw the graph having a continous drawing. That is, imagine we take a pencil and start on one point on the graph on the left side. We can draw the whole graph on the left side, but we cannot draw the graph on the right side without lifting the pencil up. As we have to "lift the pencil up" this means that the graph is not continous
Finally note that as we have a vertical asymptote at x=1 and horizontal asymptote at y=1 we have that when y is 1 or x is 1, the function y=f(x)/g(x) is undefined
Which points sre vertices of the pre-image, rectangle ABCD?Makes no sense
Given rectangle A'B'C'D', you know that it was obtained after translating rectangle ABCD using this rule:
[tex]T_{-4,3}(x,y)[/tex]That indicates that each point of rectangle ABCD was translating 4 units to the left and 3 units up, in order to obtain rectangle A'B'C'D'.
Notice that the coordinates of the vertices of rectangle A'B'C'D' are:
[tex]\begin{gathered} A^{\prime}(-5,4) \\ B^{\prime}(3,4) \\ C^{\prime}(3,1) \\ D^{\prime}(-5,1) \end{gathered}[/tex]Therefore, in order to find the coordinates of ABCD, you can add 4 units to the x-coordinate of each point and subtract 3 units to each y-coordinate of each point. You get:
[tex]\begin{gathered} A=(-5+4,4-3)=(-1,1) \\ B=^(3+4,4-3)=(7,1) \\ C=(3+4,1-3)=(7,-2) \\ D=(-5+4,1-3)=(-1,-2) \end{gathered}[/tex]Hence, the answers are:
- First option.
- Second option.
- Fourth option.
- Fifth option.
What is the value of x? A pair of intersecting lines is shown. The angle above the point of intersection is labeled left parenthesis 7 x minus 8 right parenthesis degrees. The angle directly opposite below the point of intersection is labeled left parenthesis 6 x plus 11 right parenthesis degrees. (1 point)
A –19
B 125
C 19
D 55
The value of x in the angles is 19.
How to find angles in intersecting lines?When lines intersect, angle relationships are formed such as vertically opposite angles, adjacent angles etc.
Therefore, let's find the value of x in the intersecting lines.
Hence,
7x - 8 = 6x + 11 (vertically opposite angles)
Vertically opposite angles are congruent and they share the same vertex point.
Hence,
7x - 8 = 6x + 11
subtract 6x from both sides of the equation
7x - 8 = 6x + 11
7x - 6x - 8 = 6x - 6x + 11
x - 8 = 11
add 8 to both sides of the equation
x - 8 + 8 = 11 + 8
x = 19
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What is the solution to the equation below? 6x= x + 20 O A. x = 4 B. X = 20 C. x = 5 D. No Solutions
Simplify the equation 6x = x +20 to obtain the value of x.
[tex]\begin{gathered} 6x=x+20 \\ 6x-x=20 \\ 5x=20 \\ x=\frac{20}{5} \\ =4 \end{gathered}[/tex]So answer is x = 4
Option A is correct.
For the diagram below, if < 4 = 4x - 2, and < 6 = 2x + 14, what is the value of x?Select one:a.8b.16c.4d.5
x = 8
ExplanationsFrom the line geometry shown, the line a and b are parallel lines while line "t" is the transversal.
Since the horizontal lines are parallel, hence;
[tex]\angle4=\angle6(alternate\text{ exterior angle})[/tex]Given the following parameters
[tex]\begin{gathered} \angle4=4x-2 \\ \angle6=2x+14 \end{gathered}[/tex]Equate both expressions to have:
[tex]\begin{gathered} 4x-2=2x+14 \\ 4x-2x=14+2 \\ 2x=16 \\ x=\frac{16}{2} \\ x=8 \end{gathered}[/tex]Hence the value of x is 8
Jo-o/checkpoint scatter plotsA2018161412Paw size (centimeters)10642X1b2030405066708090100Height (centimeters)Does this scatter plot show a positive association, a negative association, or no association?positive associationnegative associationno association
A scatter plot shows the association between two variables.
If the variables tend to increase and decrease together, the association is positive. If one variable tends to increase as the other decreases, the association is negative. If there is no pattern, the association is zero.
From the graph we notice that in this case both variables increcase together, therefore the scatter plot has a positive association.
what is the image of 2,10 after a dilation by a scale factor of 1/2 centered at the origin
A dilation is given by:
[tex](x,y)\rightarrow(kx.ky)[/tex]where k is the scale factor.
In this case we have:
[tex](2,10)\rightarrow(\frac{1}{2}\cdot2,\frac{1}{2}\cdot10)=(1,5)[/tex]Therefore the image is the point:
[tex](1,5)[/tex]Solving triangles using the law of cosines . Find m
The law of cosines is defined as follows:
[tex]a^2=b^2+c^2-2bc\cos A[/tex]For the given triangle
a=AC=8
b=AB=14
c=BC=11
∠A=∠B=?
-Replace the lengths of the sides on the expression
[tex]8^2=14^2+11^2-2\cdot14\cdot11\cdot\cos B[/tex]-Solve the exponents and the multiplication
[tex]\begin{gathered} 64=196+121-308\cos B \\ 64=317-308\cos B \end{gathered}[/tex]-Pass 317 to the left side of the expression by applying the opposite operation to both sides of it
[tex]\begin{gathered} 64-317=317-317-308\cos B \\ -253=-308\cos B \end{gathered}[/tex]-Divide both sides by -308
[tex]\begin{gathered} -\frac{253}{-308}=-\frac{308\cos B}{-308} \\ \frac{23}{28}=\cos B \end{gathered}[/tex]-Apply the inverse cosine to both sides of the expression to determine the measure of ∠B
[tex]\begin{gathered} \cos ^{-1}\frac{23}{28}=\cos ^{-1}(\cos B) \\ 34.77º=B \end{gathered}[/tex]The measure of ∠B is 34.77º
si f(x) = x + 5 cuanto es f(2) f(1) f(0) f(-1) f-(-2) f(a)
f (x)= x+ 5
f(2)
Reemplaza x por 2 y resuelve
f(2)= 2 + 5 = 7
Mismo procedimiento para los demas valores:
f(1) = 1 + 5 = 6
f(0) = 0 + 5 = 5
f(-1)= -1+5 = 4
f(-2)= -2+5 = 3
f(a)= a + 5
The table shows the fraction of students from differentgrade levels who are in favor of adding new items tothe lunch menu at their school. Which list shows the grade levels in order from the greatest fraction of students to the least fraction of students ?
First, write all the fractions using the same denominator. To do so, find the least common multiple of all denominatos. The denominators are:
[tex]50,20,25,75,5[/tex]The least common multiple of all those numbers is 300.
Use 300 as a common denominator for all fractions to be able to compare their values.
5th grade
[tex]\frac{33}{50}=\frac{33\times6}{50\times6}=\frac{198}{300}[/tex]6th grade
[tex]\frac{13}{20}=\frac{13\times15}{20\times15}=\frac{195}{300}[/tex]7th grade
[tex]\frac{18}{25}=\frac{18\times12}{25\times12}=\frac{216}{300}[/tex]8th grade
[tex]\frac{51}{75}=\frac{51\times4}{75\times4}=\frac{204}{300}[/tex]9th grade
[tex]\frac{3}{5}=\frac{3\times60}{5\times60}=\frac{180}{300}[/tex]Now, we can compare the numerators to list the fraction from greatest to lowest:
[tex]\begin{gathered} \frac{216}{300}>\frac{204}{300}>\frac{198}{300}>\frac{195}{300}>\frac{180}{300} \\ \Leftrightarrow\frac{18}{25}>\frac{51}{75}>\frac{33}{50}>\frac{13}{20}>\frac{3}{5} \\ \Leftrightarrow7th\text{ grade}>8th\text{ grade}>5th\text{ grade}>6th\text{ grade}>9th\text{ grade} \end{gathered}[/tex]Therefore, the list of grade levels in order from the greatest fraction of students to the least fraction of students, is:
7th grade (18/25)
8th grade (51/75)
5th grade (33/50)
6th grade (13/20)
9th grade (3/5)
Calculate the amount of money that was loaned at 4.00% per annum for 2 years if the simple interest charged was $1,240.00.
Given:-
Simple intrest is $1240. Rate is 4.00%. Time is 2 years.
To find:-
The principal amount.
The formula which relates Simple intrest, Rate, Time and Principal amount is,
[tex]I=prt[/tex]So from this the formula for p is,
[tex]p=\frac{I}{rt}[/tex]Subsituting the known values. we get,
[tex]\begin{gathered} p=\frac{I}{rt} \\ p=\frac{1240}{0.04\times2} \\ p=\frac{1240}{0.08} \\ p=\frac{124000}{8} \end{gathered}[/tex]By simplifying the above equation. we get the value of p,
[tex]\begin{gathered} p=\frac{124000}{8} \\ p=\frac{31000}{2} \\ p=15500 \end{gathered}[/tex]So the principle amount value is 15500.
3/4 square foot in 1/2 hour what is the unit rate as mixed number
Answer:
[tex]\text{1 }\frac{1}{2}[/tex]Explanation:
The unit rate is:
3/4 divided by 1/2
[tex]\begin{gathered} \frac{3}{4}\times\frac{2}{1} \\ \\ =\frac{3}{2} \end{gathered}[/tex]As a mixed fraction, it is
[tex]\text{1 }\frac{1}{2}[/tex]Kiran is solving 2x-3/x-1=2/x(x-1) for x, and he uses these steps.He checks his answer and finds that it isn’t a solution to the original equation, so he writes “no solutions.” Unfortunately, Kiran made a mistake while solving. Find his error and calculate the actual solution(s).
Solution:
Given:
[tex]\begin{gathered} To\text{ solve,} \\ \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \end{gathered}[/tex]Kiran multiplied the left-hand side of the equation by (x-1) and multiplied the right-hand side of the equation by x(x-1).
That was where he made the mistake. He ought to have multiplied both sides with the same quantity (Lowest Common Denominator) so as not to change the actual value of the question.
Multiplying both sides by the same quantity does not change the real magnitude of the question.
The actual solution goes thus,
[tex]\begin{gathered} \frac{2x-3}{x-1}=\frac{2}{x(x-1)} \\ \text{Multiplying both sides of the equation by the LCD,} \\ \text{The LCD is x(x-1)} \\ x(x-1)(\frac{2x-3}{x-1})=x(x-1)(\frac{2}{x(x-1)}) \\ x(2x-3)=2 \\ \text{Expanding the bracket,} \\ 2x^2-3x=2 \\ \text{Collecting all the terms to one side to make it a quadratic equation,} \\ 2x^2-3x-2=0 \end{gathered}[/tex]Solving the quadratic equation;
[tex]\begin{gathered} 2x^2-3x-2=0 \\ 2x^2-4x+x-2=0 \\ \text{Factorizing the equation,} \\ 2x(x-2)+1(x-2)=0 \\ (2x+1)(x-2)=0 \\ 2x+1=0 \\ 2x=0-1 \\ 2x=-1 \\ \text{Dividing both sides by 2,} \\ x=-\frac{1}{2} \\ \\ \\ OR \\ x-2=0 \\ x=0+2 \\ x=2 \end{gathered}[/tex]Therefore, the actual solutions to the expression are;
[tex]\begin{gathered} x=-\frac{1}{2} \\ \\ OR \\ \\ x=2 \end{gathered}[/tex]For an outdoor concert by the city orchestra, concert organizers estimate that 11,000 people will attend if it is not raining. If it is raining, concert organizers estimatethat 7000 people will attend. On the day of the concert, meteorologists predict a 60% chance of rain. Determine the expected number of people who will attend thisconcert
Step 1
Given;
For an outdoor concert by the city orchestra, concert organizers estimate that 11,000 people will attend if it's not raining.
If it is raining, concert organizers estimate 7000 people will attend.
On the day of the concert, meteorologists predict a 60% chance of rain.
Step 2
Given that the probability of having rain is 60%
[tex]Pr(rain)=\frac{60}{100}=0.6[/tex]So the probability of not having rain is;
[tex]\begin{gathered} Pr(rain)+Pr(no\text{ rain\rparen=1} \\ Pr(no\text{ rain\rparen=1-Pr\lparen rain\rparen} \\ Pr(no\text{ rain\rparen=1-0.6=0.4} \end{gathered}[/tex]Step 3
Now, the expected number of people who will attend the concert will be:
=(probability of not having rain x number of expected guests when it does not rain) + (probability of having rain x number of expected guests when rains)
[tex]\begin{gathered} Pr(expected\text{ number of peope\rparen=\lparen0.4}\times11000)+(0.6\times7000) \\ Pr(expected\text{ number of peope\rparen=4400+4200=8600} \end{gathered}[/tex]Answer; So, the expected number of people who will attend the concert is 8600