The functions that models the number of adopted dogs and cat is T = 3n + 7.
If the trend continues, the number of cats and dog that will be adopted by 2013 is 4600.
How to find the function that models a problem?From 1999 to 2009, the number of dogs D and the number of cats C (in hundreds) adopted from animal shelters in the United States are modelled by the equations D = 2n + 3 and C = n + 4, where n is the number of years since 1999.
Therefore, the functions that models the total number T of the adopted dogs and cats in hundreds for that time period can be represented as follows:
T = D + C
where
D = 2n + 3
C = n + 4
where
n = number of yearsT = 2n + 3 + n + 4
T = 3n + 7
b. If the trends continues the number of cats and dogs that will be adopted in 2013 can be calculated as follows:
n = 2013 - 1999 = 13Hence,
T = 3(13) + 7
T = 39 + 7
T = 46
Recall it's represented in hundred's
Therefore, 4600 dogs and cat will be adopted by 2013
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4-10x = 3+5x subtract 4 from both sides
S={1/15}
1) Solving that expression
4-10x = 3+5x Subtract 4 from both sides
4-4-10x=3-4+5x
-10x =-1+5x Subtract 5x from both sides, to isolate x on the left side
-10x -5x = -1 +5x -5x
-15x=-1 Divide both sides by -15 to get the value of x, not -15x
x=1/15
S={1/15}
You spin the spinner once. What is P(2 or odd)?
Answer:
P(2 or odd)=1
Explanation:
The spinner has 3 parts.
The probability of spinning a 2:
[tex]P(2)=\frac{1}{3}[/tex]The probability of spinning an odd number (1, 3):
[tex]P(\text{odd)}=\frac{2}{3}[/tex]Therefore:
[tex]\begin{gathered} P(2\text{ or odd)=}\frac{1}{3}+\frac{2}{3} \\ =\frac{3}{3} \\ =1 \end{gathered}[/tex]Suppose a mutual fund yielded a return of 14% last year. Its CAPM beta (β) is 1.2. The risk-free rate was 5% last year and the stock market return was 10% last year. What is the alpha (α) of the mutual fund?
The Jensen's Alpha of the mutual fund is given as follows:
α = 3.
Jensen's AlphaThe Jensen's Alpha of a mutual fund is calculated according to the rule presented as follows:
α = [Rp - (Rf + Bp x (Rm - Rf))]
The parameters of the problem are defined as follows:
Rp is the expected portfolio return.Rf is the risk free rate.Bp is the beta of the portfolio.Rm is the expected market return.Hence, in the context of this problem, the values of the parameters are given as follows:
Rp = 14, Rf = 5, Bp = 1.2, Rm = 10.
Hence the Jensen's Alpha of the mutual fund is given as follows:
α = [Rp - (Rf + Bp x (Rm - Rf))]
α = [14 - (5 + 1.2 x (10 - 5))]
α = 3.
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decide whether the events are independent or dependent and explain your answer.-drawing a ball from a lottery machine, not replacing it, and then drawing a second ball.
If the probability of an event is unaffected by other events, it is called an independent event. If the probability of an event is affected by other events, then it is called a dependent event.
A ball is drawn from a lottery machine. Then, a second ball is drawn without replacing the first ball. Let T be the number of balls in the lottery machine initially. Before the first ball is drawn, the number of balls in the machine is T. At the time the second ball is drawn, the number of balls in the machine is T-1. From T-1 balls, the second ball is drawn. So, the event of drawing the second ball is affected by the event of drawing the first ball.
Therefore, the event of drawing a ball from a lottery machine, not replacing it, and then drawing a second ball is a dependent event.
Find the distance between the following points using the pythagorean theorem (5,10) and (10,12)
Answer:
\sqrt[29]
Explanation:
Given the coordinate (5,10) and (10, 12). The formula for calculating the distance between two points is expressed as;
[tex]D\text{ =}\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}^{}[/tex]Given that;
x1 = 5
y1 = 10
x2 = 10
y2 = 12
Substitute:
[tex]\begin{gathered} D\text{ = }\sqrt[]{(10-5)^2+(12-10)^2} \\ D=\text{ }\sqrt[]{5^2+2^2} \\ D\text{ =}\sqrt[]{25+4} \\ D\text{ =}\sqrt[]{29} \end{gathered}[/tex]Hence the distance between the points is \sqrt[29]
help with this functions and equations question. please answer correctly
The distance D(t) Maya travels in her racecar and the times taken, given in the table indicates the average rate of change of distance over the specified times are;
(a) 30.3 meters per second
(b) 25.4 meters per second
What is the average rate of change of a function?The average rate of change of a function, over an interval, gives the rate at which the function changes per unit of the interval.
The average rate of change of the distance is given by the equation;
[tex] \displaystyle {Average \: rate \: of \: change = \frac{The \: sum \: of \: distance \: traveled }{The \: sum \: of \: the \: time taken } }[/tex]
The following values are obtained from the given table;
(a) At time t = 0 seconds, distance traveled, D(0) = 0 meters
At time t = 5 seconds, distance traveled, D(5) = 151.5 meters
Which gives the average rate of change as follows;
[tex] \displaystyle {Average \: rate \: of \: change = \frac{(151.5 - 0) \: m }{(5 - 0 ) \: s} = 30.3 \: m/s }[/tex]
The average rate of change for distance driven is 30.3 meters per second(b) The table gives that at time, t = 7 seconds, distance traveled, D(7) = 205.1 meters and that at time t = 9 seconds, distance traveled, D(9) = 255.9 meters, which gives;
[tex] \displaystyle {Average \: rate \: of \: change = \frac{(255.9 - 205.1) \: m }{(9 - 7) \: s} = 25.4 \: m/s }[/tex]
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what is 12 + 0.2 + 0.006 as a decimal and word form
twelve and two hundred six thousandths
On a circle of radius 9 feet, what angle would subtend an arc of length 7 feet?
________ degrees
The angle subtend an arc length of 7 feet is 44.56°
Given,
Radius of a circle = 9 feet
Arc length of a circle = 7 feet
Arc length :
The distance between two places along a segment of a curve is known as the arc length.
Formula for arc length:
AL = 2πr (C/360)
Where,
r is the radius of the circle
C is the central angle in degrees
Now,
AL = 2πr (C/360)
7 = 2 × π × 9 (C/360)
7 = 18 π (C/360)
7/18π = C/360
C = (7 × 360) / (18 × π)
C = (7 × 20) / π
C = 140 / π
C = 44.56°
That is,
The angle subtend an arc length of 7 feet is 44.56°
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Write the phrase "8 more than 10 divided by x is 12" as a variable expression:
Answer:
10/x + 8 = 12
Step-by-step explanation:
10 divided by x = 10/x
8 more than 10 divided by x = 10/x + 8
Suppose that the functions f and g are defined as follows. f(x)= x-6/x+5 g(x)= x/x+5. find f/g. Then, give its domain using an interval or union of intervals. simplify your answers.
STEP 1:
To find f/g we divide f(x) by g(x)
[tex]\frac{f}{g}=\frac{\frac{x-6}{x+5}}{\frac{x}{x+5}}\text{ = }\frac{x-6}{x+5}\text{ }\times\text{ }\frac{x+5}{x}\text{ =}\frac{x-6}{x}[/tex]Therefore the value of f/g is
[tex]\frac{f}{g}=\frac{x-6}{x}[/tex]STEP 2:
Also, the domain is the set of all possible x-values which will make the function "work", and will output real values.
The domain of this function is
[tex]-\inftyThis implies that the function would exist for all values of x except when x=0The above domain can also be represented as :
[tex](-\infty,0)\text{ and (0,}\infty)[/tex]△VWY is equilateral, VZ≅WX, and ∠XWY≅∠YVZ. Complete the proof that △VYZ≅△WYX.VWXYZ
The statement
[tex]VY\cong WX[/tex]is true because
[tex]\Delta VWY[/tex]is an equilateral triangle.
Now, the last statement is true because the triangles have 2 sides and one angle congruent, therefore, by the SAS criterion, the triangles are congruent.
Answer:4.- Triangle VWY is an equilateral triangle.
5.- SAS criterion.
5. (a) The table below shows the cumulative frequency distribution of the weight of 80 deer recorded by the zookeeper. Weight, w kg Cumulative Frequency 6 15 61-80 36 niger 81-100 58 y Determine the upper class boundary for the class 21-40. Determine the class width for the class 41-60. How many deer were recorded in the class 81-100. (iv) A deer was chosen at random from the 80 deer. What is the probability that the weight of the deer is more than 100.5 kg. Leave your answer as an EXACT value. [2]
STEP - BY - STEP EXPLANATION
What to find?
• The upper class boundary for the class 21 - 40
,• Class width for 41 - 60
,• The number of deer recorded in the class 81 - 100
Given:
(i) To find the class boundary for the class 21 - 40, we will first subtract 0.5 from 21 and then add 0.5 to 40.
That is;
20.5 - 40.5
Hence, the upper class boundary is 40.5
(ii) The class width for the class 41 - 60
The class width can be determine by subtracting 41 from 60.
That is;
[tex]60-41=19[/tex]Hence, class width = 19
(iii) Number of deer recorded in the class 81 - 100
This can be obtain by subtracting the cumulative frequency in the class from the cumulative frequency before it.
58 - 36 =22
Hence, we have 22 numbers of deer in the class 81 - 100.
(iv) A deer was chosen at random from the 80 deer. What is the probability that the weight of the deer is more than 100.5 kg.
We can solve this by first determining the number of deer that are above 100.5 kg.
Number of beer above 100.5 kg = 15 + 7 = 22
Total number of deer = 80
[tex]Probability=\frac{required\text{ outcome}}{all\text{ possible outcome}}[/tex][tex]=\frac{22}{80}[/tex][tex]=\frac{11}{40}[/tex]ANSWER
(i) 40.5
(ii) 19
(iii) 22
(iv) 11/40
Find the probability of obtaining exactly seven tails when flipping seven coins. Express your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth.
Answer:
Concept:
If you flip a coin once, there are
[tex]\text{2 possiblities}[/tex]Using the binomial probability formula below, we will have
[tex]P(x)=^nC_rp^xq^{x-r}[/tex]Where
[tex]\begin{gathered} p=probability\text{ of success} \\ q=probability\text{ of failure} \end{gathered}[/tex][tex]\begin{gathered} p=\frac{1}{2} \\ q=\frac{1}{2} \\ n=7 \\ x=7 \end{gathered}[/tex]By substituting the values, we will have
[tex]\begin{gathered} P(x)=^nC_rp^xq^{x-r} \\ P(x=7)=^7C_7(\frac{1}{2})^7(\frac{1}{2})^{7-7} \\ P(x=7)=(\frac{1}{2})^7 \\ P(x=7)=\frac{1}{128} \end{gathered}[/tex]Hence,
The final answer is
[tex]\Rightarrow\frac{1}{128}[/tex]If Lanny spins the spinner below 70 times, how many times can he expect is to land on a number divisible by 3? *
From 1 to 10, there are 3, 6, and 9 are divisible by 3
Then we have 3 choices out of 10 numbers
Since the probability = an event/outcomes
Since the event is 3
Since the outcomes are 10, then
[tex]P(\frac{no}{3})=\frac{3}{10}[/tex]This is the probability for spinning the spinner one time
But we need to spin it 70 times
We will multiply 3/10 by itself 70 times, which means make it to the power of 70
[tex]P(\frac{no}{3})=(\frac{3}{10})^{70}[/tex]The answer is (3/10)^70 OR (0.3)^70
The System of PolynomialsYou are aware of the different types of numbers: natural numbers, integers, rational numbers, and real numbers. Now you will work with a property of the number system called the closure property. A set of numbers is closed for a specific mathematical operation if you can perform the operation on any two elements in the set and always get a result that is an element of the set.Consider the set of natural numbers. When you add two natural numbers, you will always get a natural number. For example, 3 + 4 = 7. So, the set of natural numbers is said to be closed under the operation of addition.Similarly, adding two integers or two rational numbers or two real numbers always produces an integer, or rational number, or a real number, respectively. So, all the systems of numbers are closed under the operation of addition.Think of polynomials as a system. For each of the following operations, determine whether the system is closed under the operation. In each case, explain why it is closed or provide an example showing that it isn’t.1)AdditionType your response here:2)SubtractionType your response here:3)MultiplicationType your response here:4)DivisionType your response here:5)Determine whether the systems of natural numbers, integers, rational numbers, irrational numbers, and real numbers are closed or not closed for addition, subtraction, multiplication, and division.Type your response here: 6)Addition Subtraction Multiplication Division natural numbers integers rational numbers irrational numbers real numbers When a rational and an irrational number are added, is the sum rational or irrational? Explain.Type your response here:7)When a nonzero rational and an irrational number are multiplied, is the product rational or irrational? Explain.Type your response here:8)Which system of numbers is most similar to the system of polynomials?Type your response here:9)For each of the operations—addition, subtraction, multiplication, and division—determine whether the set of polynomials of order 0 or 1 is closed or not closed. Consider any two polynomials of degree 0 or 1.Type your response here:10)Polynomial 1 Polynomial 2 Operation Expression Result Degree of Resultant Polynomial Conclusion addition subtraction multiplication division What operations would the set of quadratics be closed under? For each operation, explain why it is closed or provide an example showing that it isn’t.Type your response here:11)Is there a set of expressions that would be closed under all four operations? Explain.Type your response here:
The Solution To Question Number 10:
The question says what operations would the set of quadratics be closed under.
Let the sets of quadratics be
[tex]\begin{gathered} p(x)=ax^2+bx+c \\ q(x)=mx^2+nx+k \end{gathered}[/tex]The set of two quadratics (polynomials) is closed under Addition.
Explanation:
[tex]\begin{gathered} P(x)+q(x)=(ax^2+bx+c)+(mx^2+nx+k) \\ =(a+m)x^2+(b+n)x+(c+k) \\ \text{which is still a quadratic.} \\ \text{Hence, the set of quadratics is closed under Addition.} \end{gathered}[/tex]The set of two quadratics is closed under Subtraction.
[tex]\begin{gathered} P(x)-q(x)=(ax^2+bx+c)-(mx^2+nx+k) \\ =(a-m)x^2+(b-n)x+(c-k) \\ \text{which is still a quadratic, provided both a}\ne m,\text{ b}\ne n\text{ } \\ \text{Hence, the set of quadratics is closed under Subtraction.} \end{gathered}[/tex]The set of quadratics is not closed under Multiplication.
[tex]\begin{gathered} P(x)\text{.q(x)}=(ax^2+bx+c)(mx^2+nx+k)=amx^4+(bn+ak)x^2+ck+\cdots \\ \text{Which is not a quadratic.} \\ \text{Hence, the set of quadratics is not closed under multiplication.} \end{gathered}[/tex]The set of quadratics is not closed under Division.
[tex]\begin{gathered} \text{Let the sets be f(x)=8x}^2\text{ and} \\ h(x)=2x^2-1 \\ \text{ So,} \\ \frac{f(x)}{h(x)}=\frac{8x^2}{2x^2_{}-1} \\ \text{Which is not a quadratic.} \\ \text{Hence, the set is not closed under Division.} \end{gathered}[/tex]
Aaquib can buy 25 liters of regular gasoline for $58.98 or 25 liters of permimum gasoline for 69.73. How much greater is the cost for 1 liter of premimum gasolinz? Round your quotient to nearest hundredth. show your work :)
The cost for 1 liter of premium gasoline is $0.43 greater than the regular gasoline.
What is Cost?This is referred to as the total amount of money and resources which are used by companies in other to produce a good or service.
In this scenario, we were given 25 liters of regular gasoline for $58.98 or 25 liters of premium gasoline for $69.73.
Cost per litre of premium gasoline is = $69.73 / 25 = $2.79.
Cost per litre of regular gasoline is = $58.98/ 25 = $2.36.
The difference is however $2.79 - $2.36 = $0.43.
Therefore the cost for 1 liter of premimum gasoline is $0.43 greater than the regular gasoline.
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Please help with the question below (please try to answer in maximum 10/15 minutes).
Solution:
Given the dimensions of the composite figure below
[tex]\begin{gathered} For\text{ the cuboid:} \\ l=12\text{cm} \\ w=4\text{ cm} \\ h=3cm \\ For\text{ the triangular prism:} \\ a=3\text{ cm} \\ b=4\text{ cm} \\ c=13\text{ cm} \\ h=5\text{ cm} \end{gathered}[/tex]To find the surface area, SA, of the composite figure, the formula
[tex]SA=2(lh)+2(wh)+(lw)+2(\frac{1}{2}lh)+(bc)+(ah)[/tex]
Substitute the values of the variables into the formula above
[tex]\begin{gathered} SA=2\left(12\cdot3\right)+2\left(3\cdot4\right)+\left(12\cdot4\right)+2\left(\frac{1}{2}\left(12\cdot5\right)\right)+\left(13\cdot4\right)+\left(4\cdot5\right) \\ SA=2(36)+2(12)+(48)+(60)+(52)+20 \\ SA=72+24+48+60+52+20 \\ SA=276\text{ cm}^2 \end{gathered}[/tex]Hence, the surface area, SA, is
[tex]276\text{ cm}^2[/tex]please help me understand how to find the average rate of change of the function over the given interval and please show me work.
To answer this, you'll need to recall a formula for finding the rate of change of one variable with respect to another. Given f(x)=x^2 + x +1, the rate of change of the variable with respect to x is given by:
[tex]\begin{gathered} \frac{\differentialD yy}{\square}y}{dx}=n(ax^{n-1}),\text{ where n is the power of variable term, and a is the coefficient.}y}{\square}yy}{dx}=\text{nax}^{n-1} \\ So\text{ when f(x)=x\textasciicircum 2+x+1 is differentiated, we will arrive at } \\ \\ \frac{dy}{dx}=2x+1\text{ The average rate of change of the function within the range (-3,-2) means, we have to use x as -3 and also x as -2 into the derivative function } \\ x=-3 \\ \frac{\differentialD yy}{\square}y}{dx}=2(-3)+1=-6+1=-5y}{\square}y}{dx}=2(-3)+1=-6+1=-5 \\ \text{Also, } \\ x=-2 \\ \frac{\differentialD yy}{\square}y}{dx}=2x+1\text{ becomes}y}{\square}yy}{dx} \\ \\ \end{gathered}[/tex]What is the slope of a line parallel to the line whose equation is 5x-3y=18
Slope and slant both refer to an incline away from a reference surface or line that is generally straight.The definition of slope is "a vertical inclination in an oblique direction"Here, the land abruptly slopes either upward or downhill.
How to Determine a Line's Slope?
slope,The inclination of a line with respect to the horizontal is measured numerically.The ratio of the vertical to the horizontal distance between any two points on a line, ray, or line segment is known as its slope in analytic geometry ("slope equals rise over run"). To determine how much the y coordinates have changed, find the difference.To determine how much the x coordinates have changed, find the difference.Find the slope by dividing y by x.
Y=mx+b, where m is the slope and b is the y-intercept, is the slope-intercept form.
y=mx+b
Change the formula to 3y+18=5x.
−3y+18=5x
From both sides of the equation, deduct 18.
−3y=5x−18
Simplify by multiplying each term in 3y=5x18 by 3.
Subtract 3 from each term in 3y=5x18.
−3y/−3=5x/−3+−18/−3
Make the left side simpler.
Tap to take additional steps:Y= -5X/3+-18/-3
Make the right side simpler.
Tap to take additional steps: y=5x/3+6
Write in the form y=mx+b.
Tap to take additional steps: y=5/3x+6
The slope is- 5/3 using the slope-intercept form.
.
m=−5/3
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Sean johnson 1. The angles of a triangle are described as follows: angle A is the largest angle: its measure is twice the measure of angle B. The measure of angle C is 2 less than half the measure of angle B. Find the measures of the three angles in degrees.
Explanation
Let
angle A is the largest angle
[tex]m\measuredangle A=m\measuredangle A[/tex]its measure is twice the measure of angle B,( in other words you have to multiply angle b by 2, to get angle A)
[tex]2(m\measuredangle B)=m\measuredangle A[/tex]Find the smallest distinct positive numbers that provide a counterexample to show the statement is false.The sum of any two different odd numbers plus any even number is odd.
The sum of two even or odd numbers ALWAYS gives an even number.
We'll run a test with 1,2 and 3.
Odd numbers: 1, 3
Even number: 2
Adding the odd numbers, we get 1 +3 = 4.
Adding it to the even number, we get 4 +2 = 1 + 3 + 2 =6
The general form of an odd number = 2n + 1
The general form of an even number = 2n
Adding 2 odd numbers give 2(2n + 1) = 4n + 2
Adding to an even number; 4n + 2 + 2n
Giving 6n + 2
Any number of the form above is an even number
The statement is thus false.
Help with these two questions please. Match the sentence with a word
EXPLANATION
Given that two angles form a linear pair, we can assevere that the postulate that applies is the Linear Pair Postulate.
Find 5 number summary for data given
The 5 number summary of the data given is:
Minimum = 59
Q1 = 66.50
Median = 78
Q3 = 90
Maximum = 99
What is the 5 number summary?A stem and leaf plot is a table that is used to display a dataset. A stem and leaf plot divides a number into a stem and a leaf. The stem is the first digit in a number while the leaf is the second digit in the number.
The minimum is the smallest number in the stem and leaf plot. This is 59. Q1 is the first quartile.
Q1 = 1/4 x (n + 1)
Where n is the total number in the dataset
1/4 x 19 = 4.75 term
(64 + 69) / 2 = 66.50
Q3 is the third quartile.
Q1 = 3/4 x (n + 1)
Where n is the total number in the dataset
3/4 x 19 = 14.25 term = 90
The median is the number that is at the center of the dataset.
Median = 1/2(n + 1)
1/2 x 19 = 8.5 term
(76 + 80) / 2 = 78
The maximum is the largest number in the dataset. This number is 99.
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I need help with some problems on my assignment please help
The circumcenter of a triangle is the center of a circumference where the three vertex are included. So basically we must find the circumference that passes through points O, V and W. The equation of a circumference of a radius r and a central point (a,b) is:
[tex](x-a)^2+(y-b)^2=r^2[/tex]We have three points which give us three pairs of (x,y) values that we can use to build three equations for a, b and r. Using point O=(6,5) we get:
[tex](6-a)^2+(5-b)^2=r^2[/tex]Using V=(0,13) we get:
[tex](0-a)^2+(13-b)^2=r^2[/tex]And using W=(-3,0) we get:
[tex](-3-a)^2+(0-b)^2=r^2[/tex]So we have a system of three equations and we must find three variables: a, b and r. All equations have r^2 at their right side. This means that we can take the left sides and equalize them. Let's do this with the second and third equation:
[tex]\begin{gathered} (0-a)^2+(13-b)^2=(-3-a)^2+(0-b)^2 \\ a^2+(13-b)^2=(-3-a)^2+b^2 \end{gathered}[/tex]If we develop the squared terms:
[tex]a^2+b^2-26b+169=a^2+6a+9+b^2[/tex]Then we substract a^2 and b^2 from both sides:
[tex]\begin{gathered} a^2+b^2-26b+169-a^2-b^2=a^2+6a+9+b^2-a^2-b^2 \\ -26b+169=6a+9 \end{gathered}[/tex]We substract 9 from both sides:
[tex]\begin{gathered} -26b+169-9=6a+9-9 \\ -26b+160=6a \end{gathered}[/tex]And we divide by 6:
[tex]\begin{gathered} \frac{-26b+160}{6}=\frac{6a}{6} \\ a=-\frac{13}{3}b+\frac{80}{3} \end{gathered}[/tex]Now we can replace a with this expression in the first equation:
[tex]\begin{gathered} (6-a)^2+(5-b)^2=r^2 \\ (6-(-\frac{13}{3}b+\frac{80}{3}))^2+(5-b)^2=r^2 \\ (\frac{13}{3}b-\frac{62}{3})^2+(5-b)^2=r^2 \end{gathered}[/tex]We develop the squares:
[tex]\begin{gathered} (\frac{13}{3}b-\frac{62}{3})^2+(5-b)^2=r^2 \\ \frac{169}{9}b^2-\frac{1612}{9}b+\frac{3844}{9}+b^2-10b+25=r^2 \\ \frac{178}{9}b^2-\frac{1702}{9}b+\frac{4069}{9}=r^2 \end{gathered}[/tex]So this expression is equal to r^2. This means that is equal
Which answer choice below is a solution to this equation?7x + 5 – 2x = 2x – 7A. 2B. 0C. -4D. 8
SOLUTION:
Step 1:
In this question, we are given the following:
[tex]7x\text{ + 5 - 2 x = 2 x- 7}[/tex]Step 2:
The details of the solution are as follows:
[tex]\begin{gathered} 7\text{ x - 5 - 2 x = 2 x - 7} \\ 5x\text{ - 5 = 2x - 7} \\ collecting\text{ like terms, we have that:} \\ 5\text{ x - 2x = - 7 - 5} \\ 3\text{ x = - 12} \\ Divide\text{ both sides by 3, we have that:} \\ x\text{ =}\frac{-12}{3} \\ x\text{ = - 4 \lparen OPTION C \rparen} \end{gathered}[/tex]CONCLUSION:
The final answer is:
[tex]x\text{ = - 4 \lparen OPTION C\rparen}[/tex]
Blue whales can weigh as much as 150 tons. Convert the weight to pounds.
SOLUTION:
The conversion formula from tons to pounds is;
[tex]1\text{ }US\text{ }ton=2000\text{ }pounds[/tex]Thus, converting this to pounds, the Blue whale would weigh;
[tex]150\times2000=300,000\text{ }pounds[/tex]Thus, the whale weighs 300,000 pounds
Answer:
The answer is C: 150/y
Step-by-step explanation:
Susan is flying a kite, which gets caught in the top of a tree. Use the diagram to estimate the height of the tree. a. 87 ft b. 74 ft c. 65 ft d. 63 ft
Given the information on the picture, we have the following right triangle:
we can use the tangent trigonometric function to find the height of the tree:
[tex]\begin{gathered} tan(44)=\frac{\text{opposite side}}{adjacent\text{ side}}=\frac{h}{90} \\ \Rightarrow\tan (44)=\frac{h}{90} \end{gathered}[/tex]solving for h, we get:
[tex]\begin{gathered} \frac{h}{90}=\tan (44) \\ \Rightarrow h=90\cdot\tan (44)=86.9\approx87 \\ h=87ft \end{gathered}[/tex]therefore, the height of the tree is 87 ft
Madeline is a salesperson who sells computers at an electronics store. She makes a base pay of $80 each day and then is paid a $20 commission for every computer sale she makes. Make a table of values and then write an equation for P,P, in terms of x,x, representing Madeline's total pay on a day on which she sells xx computers.
I need equation
The equation for 'P', representing Madeline's total pay on a day on which she sells 'x' computers is → P = 80 + 20x.
Given, At an electronics store, Madeline sells computers as a salesperson. She receives a $80-per-day base salary in addition to a $20 commission for each computer she sells.
What is Equation Modelling?
Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem.
We can model the equation for Madeline's total pay as follows -
P = base pay + (number of sold computer) × (cost of 1 computer)
P = 80 + 20x
Therefore, the equation for 'P', representing Madeline's total pay on a day on which she sells 'x' computers is → P = 80 + 20x
Learn more about Equation Modelling here:
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NO LINKS!! Please help me with this probability question. 4a
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Explanation:
mu = 500 = mean
sigma = 100 = standard deviation
We'll need the z score for x = 620
z = (x - mu)/sigma
z = (620-500)/100
z = 1.20
The task of finding P(x > 620) is equivalent to P(z > 1.20)
Use a Z table or a Z calculator to find that
P(Z < 1.20) = 0.88493
which leads to
P(Z > 1.20) = 1 - P(Z < 1.20)
P(Z > 1.20) = 1 - 0.88493
P(Z > 1.20) = 0.11507
This converts to 11.507% and rounds to 11.5%
About 11.5% of the students score higher than a 620 on the SAT.
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Another approach:
Open your favorite spreadsheet program. The command we'll be using is called NORMDIST. The template is this
NORMDIST(x, mu, sigma, 1)
x = 620 = critical valuemu = 500 = meansigma = 100 = standard deviationThe 1 at the end tells the spreadsheet to use a CDF instead of PDF. Use 0 if you want a PDF value.If you were to type in [tex]\text{=NORMDIST(620,500,100,1)}[/tex] then you'll get the area under the normal distribution to the left of x = 620
This means [tex]\text{=1-NORMDIST(620,500,100,1)}[/tex] will get us the area to the right of 620. The result of that calculation is approximately 0.11507 which leads to the same answer of 11.5% as found earlier.
When using a spreadsheet, don't forget about the equal sign up front. Otherwise, the spreadsheet will treat the input as text and won't evaluate the command.
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Another option is to use a TI83 or TI84 calculator.
Press the button labeled "2nd" in the top left corner. Then press the VARS key. Scroll down to "normalcdf"
The template is
normalcdf(L, U, mu, sigma)
L = lower boundaryU = upper boundarymu = mean sigma = standard deviationThe mu and sigma values aren't anything new here. But the L and U are. In this case L = 620 is the lower boundary and technically there isn't an upper boundary since it's infinity. Unfortunately the calculator wants a number here, so we just pick something very large. You could go for U = 99999 as the stand in for "infinity". The key is to make sure it's more than 3 standard deviations away from the mean.
So if you were to type in [tex]\text{normalcdf(620,99999,500,100)}[/tex] then the calculator will display roughly 0.11507, which is in line with the other answers mentioned earlier.
As you can see, there are many options to pick from. Searching out "normal distribution calculator" or "z calculator" will yield many free options. Feel free to pick your favorite.
Use the graph below to determine the equation of the circle in (a) center-radius form and (b) general form.10-(-3,6)(-6,3(0,3)-10(-3,0)1010
Question:
Solution:
An equation of the circle with center (h,k) and radius r is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]This is called the center-radius form of the circle equation.
Now, in this case, notice that the center of the circle is (h,k) = (-3,3) and its radius is r = 3 so that the center-radius form of the circle would be:
[tex](x+3)^2+(y-3)^2=3^2[/tex]To obtain the general form, we must solve the squares of the previous equation:
[tex](x+3)^2+(y-3)^2-3^2\text{ = 0}[/tex]this is equivalent to:
[tex](x^2+6x+3^2)+(y^2-6y+3^2)\text{ - 9 = 0}[/tex]this is equivalent to
[tex]x^2+6x+9+y^2-6y\text{ = 0}[/tex]this is equivalent to:
[tex]x^2+y^2+6x-6y\text{ +9= 0}[/tex]so that, the general form equation of the circle would be:
[tex]x^2+y^2+6x-6y\text{ +9= 0}[/tex]thus, the correct answer is:
CENTER - RADIUS FORM:
[tex](x+3)^2+(y-3)^2=3^2[/tex]GENERAL FORM:
[tex]x^2+y^2+6x-6y\text{ +9= 0}[/tex]