EXPLANATION
Since he only made 13 from 19 throws, the percent error is as shown as follows:
[tex]Percent\text{ error=}\frac{\parallel measured-real\parallel}{real}*100[/tex]Plugging in the numbers into the expression:
[tex]Percent\text{ error=}\frac{\parallel19-13\parallel}{13}*100[/tex]Subtracting numbers:
[tex]Percent\text{ error=}\frac{6}{13}*100[/tex]Multiplying terms:
[tex]Percent\text{ error=46.15\%}[/tex]In conclusion, the percent error was 46.15%
What is the rational expression as a sum of partial fractions?
Rewrite the expression as:
[tex]\frac{-x^2+2x-5}{x^2(x-1)}[/tex]The partial fraction expansion is of the form:
[tex]\frac{-x^2+2x-5}{x^2(x-1)}=\frac{A}{x-1}+\frac{B}{x}+\frac{C}{x^2}[/tex]Multiply both sides by x²(x - 1):
[tex]\begin{gathered} -x^2+2x-5=Ax^2+(x-1)(Bx+C) \\ -x^2+2x-5=-C+(A+B)x^2+(C-B)x \end{gathered}[/tex]Equate the coefficients on both sides:
[tex]\begin{gathered} -5=-C_{\text{ }}(1)_{} \\ 2=C-B_{\text{ }}(2) \\ -1=A+B_{\text{ }}(3) \end{gathered}[/tex]So, from (1):
[tex]C=5[/tex]Replace C into (2):
[tex]\begin{gathered} 2=5-B \\ B=3 \end{gathered}[/tex]Replace B into (3):
[tex]\begin{gathered} -1=A+3 \\ A=-4 \end{gathered}[/tex]Therefore, the answer is:
[tex]\frac{-x^2+2x-5}{x^2(x-1)}=\frac{-4}{x-1}+\frac{3}{x}+\frac{5}{x^2}[/tex]Are these lines parallel or not:L1 : (2,-1), (5,-7), and L2: (0,0), (-1,2) A. ParallelB.No
So,
Two lines are parallel when their slopes are the same.
So, let's find the slope of each line, and then compare them.
[tex]\begin{gathered} L_1\colon(x_1,y_1)=(2,-1);\text{ }(x_2,y_2)=(5,-7) \\ m=\frac{y_2-y_1}{x_2-x_1} \end{gathered}[/tex]Replacing the ordered pairs in the equation, we obtain:
[tex]m=\frac{-7-(-1)}{5-2}=\frac{-6}{3}=-2[/tex]Thus the slope of the first line is -2. Let's use the same process to find the slope of the second line:
[tex]L_2\colon(x_1,y_1)=(0,0);\text{ }(x_2,y_2)=(-1,2)[/tex]Given:
[tex]m=\frac{y_2-y_1}{x_2-x_1}\to m=\frac{2-0}{-1-0}=\frac{2}{-1}=-2[/tex]As you can see, the slope of both lines is the same. So, the lines are parallel.
There are 15 tables set up for a banquet, of which 3 have purple tablecloths.What is the probability that a randomly selected table will have a purple tablecloth?Write your answer as a fraction or whole number.P(purple)
To determine the probability of an event "A" you have to calculate the quotient of the number of favorable outcomes for A and the number of possible outcomes.
[tex]P(A)=\frac{nº\text{favorable outcomes}}{nº\text{ of possible outcomes}}[/tex]Let the event of interest be A: the table has a purple tablecloth.
The favorable outcomes for this event will be the number of tables that have a purple tablecloth, in this case there are 3 tables with purple tablecloth.
The number of possible outcomes is given by the total number of tables that are set up, which are 15 tables.
You can calculate the probability of A as follows:
[tex]\begin{gathered} P(A)=\frac{nº\text{ tables with purple tablecloth}}{nº\text{ tables}} \\ P(A)=\frac{3}{15}=\frac{1}{5} \end{gathered}[/tex]The probability of selecting a table at random and that it will have purple tablecloth is 1/5
sally built a small ramp for her bicycle. the ramp 10 meters long and creates a 15 degrees angles with the ground. how tall is the ramp at the highest point?
A right triangle is formed, where the length of the ramp (10 m) is the hypotenuse, and the height (h) of the ramp is the opposite side to the angle of 15°.
From defintion:
[tex]\sin (angle)=\frac{\text{opposite side}}{hypotenuse}[/tex]Substituting with data:
[tex]\begin{gathered} \sin (15)=\frac{h}{10} \\ \sin (15)\cdot10=h \\ 2.6\text{ m =h} \end{gathered}[/tex]The ramp is 2.6 meters tall
Which image describes the translation of pre - image ABCD to the transformed image A’ B’ C’ D’?A) (x, y) ➝ (x - 2, y - 3)B) (x, y) ➝ (x - 3, y - 2)C) (x, y) ➝ (x + 2, y + 3)D) (x, y) ➝ (x + 3, y + 2)
x, y)➝(x - 3, y - 2)
Option B is correct
Explanations:Step 1: Determine the coordinates of the vertices of ABCD
A(-1, 4), B(2, 3), C(1, 1), D(-3, 3)
Step 2: Determine the coordinates of the vertices of A'B'C'D'
A'(-4, 2), B'(-1, 1), C'(-2, -1), D'(-6, 1)
Step 3: Determine the translation of the x and y coordinates of ABCD to the x and y coordinates of A'B'C'D'
Note that there is a difference of 3 in the x coordinate, and a difference of 2 in the y-coordinate.
Therefore, the rule of translation is:
(x, y)➝(x - 3, y - 2)
an airplane can travel 380 miles per hour in the air if it travels 2562 miles with the wind in the same length of time it travels 1998 miles against the wind, what is the speed of the wind?
Given
380 miles per hour
2562 miles with the wind
1998 miles against
Procedure
v = speed
v = 380 miles per hour
w = wind
v1 = v + w
v2 = v - w
t = same length of time
[tex]\begin{gathered} x_1=v_1t \\ x_2=v_2t \\ \\ \frac{x_1}{v_1}=\frac{x_2}{v_2} \\ \\ x_1(v_2)=x_2v_1 \\ x_1v_2-x_2v_1=0 \\ 2562(380-w)-1998(380+w)=0 \\ 973560-2562w-759240-1998w=0 \\ 214320-4560w=0 \\ w=\frac{214320}{4560} \\ w=47 \end{gathered}[/tex]The answer would be w = 47 miles per hour
PLEASE HELP DUE VERY SOON
The six trigonometric functions of the angle θ are Sinθ = (2√14)/9 , Cosθ = 5/9 , Tanθ = (2√14)/5 , Secθ = 9/5,
Cosecθ = 9/(2√14) , Cotθ = 5/(2√14) .
In the question a right triangle is given
where for angle θ , the base(b) is 10 , the hypotnuse(h) is 18 ,
we find the perpendicular by Pythagoras Theorem
base² + perpendicular² = hypotnuse²
10² + perpendicular² = 18²
perpendicular² = √(324-100) = 4√14
perpendicular(p) = 4√14
So , Sinθ = p/h = (4√14)/18 = (2√14)/9 ,
Cosθ = b/h = 10/18 = 5/9,
Tanθ = p/b = (4√14)/10 = (2√14)/5,
Sec θ = h/b = 18/10 = 9/5,
Cosec θ = h/p = 18/(4√14) = 9/(2√14),
Cot θ = b/p = 10/(4√14) = 5/(2√14)
Therefore , the six trigonometric functions of the angle θ is Sinθ = (2√14)/9 , Cosθ = 5/9 , Tanθ = (2√14)/5 , Secθ = 9/5,
Cosecθ = 9/(2√14) , Cotθ = 5/(2√14) .
Learn more about Trigonometry here
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Which of the following is a quadratic function? O A. y= - x2 +16V - 64 x² 5-64 O B. y=2x2 - 12x+9 O c. – 3x+2y= -6 D. * = 2y2 - 9y+4
A quadratic function has the form:
y = ax^2 +bx + c
where: a, b and c are not equal to zero.
So, the correct option is B.
y= 2x^2 - 12x + 9
Dylan is driving to a concert and needs to pay for parking. There is an automatic fee of $5 just to enter the parking lot, and when he leaves the lot, he will have to pay an additional $2 for every hour he had his car in the lot. How much total money would Dylan have to pay for parking if he left his car in the lot for 6 hours? How much would Dylan have to pay if he left his car in the lot for tt hours?Cost of parking for 6 hours: Cost of parking for tt hours:
Here, we have a fixed parking cost and a variable parking lot that is a function of the time spent in the lot. Our approach is to create an algebraic relationship and then slot in our variable vales to solve.
Let c represent the cost of parking.
Let t represent the time car spent in the lot.
We then have:
[tex]c=5+2t[/tex]To calculate the cost of 6 hours in the lot, we have:
[tex]\begin{gathered} c=5+2(6) \\ c=5+12=17 \end{gathered}[/tex]$17 for a 6 hour packing.
Cost for tt hours.
[tex]c=5+2(tt)[/tex]Cost for tt hours = c = 5+2(tt)
A 7.5 % of what amount gives $37.50? ANS. $ _________.
If 7.5% of a certain amount gives $37.50
To obtain the amount
Step 1: let the unknown amount be y
7.5% of y will be:
[tex]\frac{7.5}{100}\times y=\frac{7.5y}{100}=0.075y\text{ }[/tex]Step 2: Equate 0.075y to $37.5 and then solve for y
[tex]\begin{gathered} 0.075y=37.5 \\ \text{divide both sides by 0.075} \\ \frac{0.075y}{0.075}=\frac{37.5}{0.075} \end{gathered}[/tex]Then,
[tex]y=\frac{37.5}{0.075}=500[/tex]Hence, the original amount is $500
Hello! May I please have some help on this one? I worked it out but my answer is wrong.
SOLUTION
Arranging r books from n books is given as n permutation r
[tex]\begin{gathered} ^nP_r\text{ is the number of ways of arranging r books from n books } \\ \\ n=10 \\ r=4 \\ \\ ^{10}P_4=5040\text{ \lbrack input into calculator \rbrack} \end{gathered}[/tex]Which of the following is the equation of the line of best fit?A. Y= 50/3xB. Y= 20/3xC. Y=2/3x D. Y=5/3x
The best fit line passes through the point (3,50) and (6,100).
The euation of line passing through point (x_1,y_1) and (x_2,y_2) is,
[tex]y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)[/tex]Determine the equation of line passing through point (3,50) and (6,100).
[tex]\begin{gathered} y-50=\frac{100-50}{6-3}(x-3) \\ y-50=\frac{50}{3}(x-3) \\ y=\frac{50}{3}x-\frac{50}{3}\cdot3+50 \\ y=\frac{50}{3}x \end{gathered}[/tex]So equation of best fit line is y = 50/3x.
Option A is correct answer.
Expand (y + 1)(y + 4)
The given expression is:
(y + 1) (y + 4)
To expand the expression, each of the terms in the first bracket multiplies each term in the second bracket
The expression then becomes:
[tex]\begin{gathered} y^2+\text{ 4y + y + 4} \\ y^2\text{ + 5y + 4} \end{gathered}[/tex]Evaluate the expression when b=-4 and c = 2 9c-b
Given that b = -4 and c = 2;
To evaluate 9c - b, we will substitute the given values of b and c into the expression.
Thus, we have:
[tex]\begin{gathered} 9c-b \\ 9(2)-(-4) \\ 18+4 \\ =22 \end{gathered}[/tex]can you please help me with this it's on surface area of prisms/cylinders and word problems
Solution
For this case we can find the lenghts of the slants like this:
[tex]s2=\sqrt[]{4.5^2+3.5^2}=5.70[/tex][tex]undefined[/tex]See your levelsELEWhat is the area of this figure?9 ft3 ft7 ft5 ft2 ft16 ftsquare feetSubmit
S = 59 ft²
1) Let's find out the area of that figure by tracing auxiliary lines to decompose that into smaller rectangles:
2) So we can calculate the rectangle to the left
S = w * l Rectangle Area formula
S = 7 x 2
S = 14 ft²
And the larger one:
S= 9 x 5
S = 45 ft²
3) Now let's add them up to find out the area of the polygon:
S= S_1 +S_2
S = 14 +45
S = 59 ft² And that's the answer
You place a cup of 210oF coffee on a table in a room that is 68oF, and 10 minutes later, it is 200oF. Approximately how long will it be before the coffee is 180oF? Use Newton's law of cooling:A.45 minutesB.33 minutesC.1 hourD.15 minutes
We know that Newton's law of cooling is:
[tex]\begin{gathered} T(t)=T_A+(T_O-T_A)e^{-kt} \\ \text{ where} \\ T_A\text{ is the temperature of the surrounding} \\ T_O\text{ is the temperature of the object} \\ k\text{ is the cooling constant } \end{gathered}[/tex]From the problem we know that the cup was originally at 210°F and that the room is at 68°F, then we have:
[tex]\begin{gathered} T(t)=68+(210-68)e^{-kt} \\ T(t)=68+142e^{-kt} \end{gathered}[/tex]To determine the value of k we use the fact that after 10 minutes the temperature of the object is 200°F, then we have:
[tex]\begin{gathered} 142e^{-10k}+68=200 \\ 142e^{-10k}=200-68 \\ 142e^{-10k}=132 \\ e^{-10k}=\frac{132}{142} \\ -10k=\ln(\frac{66}{71}) \\ k=-\frac{1}{10}\ln(\frac{66}{71}) \end{gathered}[/tex]Then the temperature of the cup is decreasing according to the function:
[tex]T(t)=142e^{\frac{1}{10}\ln(\frac{66}{71})t}+68[/tex]Now, that we have the function we can determine how much time it will take for the cup to be 180°F we equate our function to this temperature and solve for t:
[tex]\begin{gathered} 142e^{\frac{1}{10}\ln(\frac{66}{71})t}+68=180 \\ 142e^{\frac{1}{10}\ln(\frac{66}{71})t}=180-68 \\ e^{\frac{1}{10}\ln(\frac{66}{71})t}=\frac{112}{142} \\ \frac{1}{10}\ln(\frac{66}{71})t=\ln(\frac{56}{71}) \\ t=\frac{10\ln(\frac{56}{71})}{\ln(\frac{66}{71})} \\ t\approx32.5 \end{gathered}[/tex]Therefore, it will take approximately 33 minutes for the cup to be 180°F
The volume of the rectangular box shown below is 96. If the base of the box has the dimensions shown, what is the height of the box? 4 3
we get that the height is
[tex]h=\frac{96}{4\cdot3}=8[/tex]so we get that the height is 8
Write an equation of a line that is perpendicular to y=3x+5 and contains the point (0,0).
To write the equation of a line you need to identify the slope (m) and the y-intercept (b):
[tex]y=mx+b[/tex]Slope:
Perpendicular lines have slopes that are opposite of the reciprocal of each other.
As the given line is perpendicular to y=3x+5 (slope m=3) the slope is:
[tex]\begin{gathered} \text{slope}=m \\ \\ Perpendicular\text{ slope = -}\frac{1}{m} \\ \\ Perpendicular\text{ slope = }-\frac{1}{3} \end{gathered}[/tex]Y-intercept:
Value of the line (y coordinate) when it cross the y-axis (when x is equal to 0)
As the given line contains the point (0,0) the value of y when x is 0 is also 0
y.intercept=0
Then, you get the next equation for the line:
[tex]y=-\frac{1}{3}x[/tex]What is the best choice for the common denominator in this problem.
Given:-
[tex]\frac{1}{5}+\frac{2}{6}[/tex]To find the required value.
So to add the given fraction. first we should have same denominator. so we take LCM,
[tex]\text{LCM of 5 and 6 is 30.}[/tex]So we get,
[tex]\frac{1}{5}+\frac{2}{6}=\frac{1\times6}{5\times6}+\frac{2\times5}{6\times5}=\frac{6}{30}+\frac{10}{30}=\frac{16}{30}[/tex]So the correct denominator is 30.
Find the trigonometric ratio (NOT the angle) of cos X (write as a fraction in lowest terms.) x=19y=22z=29
Cosidering angle X in the triangle we have the adjacenb leg being 2y= 2 and the hypotenuse of the triangle i sz-= 29.
We know that
[tex]cos=\frac{adjacent}{hyp}[/tex]So:
[tex]cosX=\frac{22}{29}[/tex]Answer: CosX = 22/29
Line segment XY begins at ( - 6,4) and ends at ( - 2,4). The segment is reflected over the x-axis and translated left 3 units to form line segment X ‘ Y ‘. Enter the length , in units , of the lines segment X’ Y’ .
ANSWER
4 units
EXPLANATION
The transformations made to the line segment XY are a reflection and a translation. Both of these transformations do not change the size of the figure, so the length of line segment X'Y' is the same as the length of line segment XY.
The distance between two points (x₁, y₁) and (x₂, y₂) is found with the Pythagorean Theorem,
[tex]d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^^2}[/tex]In this case, the endpoints of line segment XY are (-6, 4) and (-2, 4), so its length is,
[tex]d=\sqrt{(-6-(-2))^2+(4-4)^2}=\sqrt{(-6+2)^2+0^2}=\sqrt{(-4)^2}=\sqrt{16}=4[/tex]Hence, the length of line segment X'Y' is 4 units.
Use the formula for present value of money to calculate the amount you need to invest now in one lump sum in order to have $1,000,000 after 40 years with an APR of 5% compounded quarterly. Round your answer to the nearest cent, if necessary.
Given:
There are given that the initial amount, time period, and rate are:
[tex]\begin{gathered} future\text{ value:1000000} \\ time\text{ period:40 year} \\ rate:\text{ 5\%} \end{gathered}[/tex]Explanation:
To find the present value, we need to use the present value formula:
So,
From the formula of present value:
[tex]PV=FV\frac{1}{(1+\frac{r}{n})^{nt}}[/tex]Then,
Put all the given values into the above formula:
So,
[tex]\begin{gathered} PV=FV\frac{1}{(1+\frac{r}{n})^{nt}} \\ PV=1000000\frac{1}{(1+\frac{0.05}{4})^{4\times40}} \end{gathered}[/tex]Then,
[tex]\begin{gathered} PV=1,000,000\times\frac{1}{(1+\frac{0.05}{4})^{4\times40}} \\ PV=1,000,000\times\frac{1}{(1.0125)^{160}} \\ PV=1,000,000\times\frac{1}{7.298} \\ PV=137023.84 \end{gathered}[/tex]Final answer:
Hence, the amount is $137023.84
Cómo hallar el valor de variables
Variables are the unknows, on this case we have 2 variables
Y and X
variables are always represented by letters
X is
we write the original equation
[tex]7y+3x=9[/tex]now we try to x be alone to solve, then we subtract 7y on both sides
[tex]\begin{gathered} (7y-7y)+3x=9-7y \\ 0+3x=9-7y \\ 3x=9-7y \end{gathered}[/tex]now we divide on both sides by 3 to solve x
[tex]\begin{gathered} \frac{3x}{3}=\frac{9-7y}{3} \\ \\ x=\frac{9-7y}{3} \end{gathered}[/tex]Y is
write original equation
[tex]7y+3x=9[/tex]subtract 3x on both sides to remove 3x on right
[tex]\begin{gathered} 7y+(3x-3x)=9-3x \\ 7y+0=9-3x \\ 7y=9-3x \end{gathered}[/tex]and divide by 7 to solve y
[tex]\begin{gathered} \frac{7y}{7}=\frac{9-3x}{7} \\ \\ x=\frac{9-3x}{7} \end{gathered}[/tex]Which statement describes the effect on the parabola y = x2 + 4x + 5 when it ischanged to y = x2 + 4x - 2?
ANSWER
EXPLANATION
We want to find the effect on the parabola when it is changed from:
[tex]y=x^2+4x+5[/tex]to
[tex]undefined[/tex]select the correct answer. and the figure, angle k measures 45°. what is the measurement of angle c? 38° 45° 90° 98°.
From the figure, we can conclude that the little triangle is an isosceles triangle, the greatest angle is 90 because ∠A = 90 and they are supplementary, therefore, using the triangle sum theorem:
[tex]\begin{gathered} m\angle J=m\angle K \\ m\angle K+m\angle K+90=180 \\ 2m\angle K=180-90 \\ 2m\angle K=90 \\ m\angle K=\frac{90}{2} \\ m\angle K=45 \end{gathered}[/tex]Please help with with wuestion 1 I really need help
Given: Evan illustrates 1/6 of a children's book of paper in 1/3 hours.
Required: To determine how long Evan will take to illustrate an entire page.
Explanation: Let x denotes the length of the complete page. Then according to the question, the time taken to illustrate 1/6th of x is 1/3 hours, i.e.
[tex]x\times\frac{1}{6}=\frac{1}{3}[/tex]We need to determine how long Evan will take to illustrate the complete page. So the time taken to illustrate x is-
[tex]\begin{gathered} x=\frac{1}{3}\times6 \\ =2\text{ hours} \end{gathered}[/tex]Final Answer: It will take Evan 2 hours to illustrate an entire page.
What is the image point of (4, -6) after a translation right 5 units and up 4 units?Submit Answer
When you translate the pre-image point (x,y) right 5 units and up 4 units, we have the image point:
[tex](x,y)\rightarrow(x+5,y+4)[/tex]Therefore, the image point of (4, -6) after a translation right 5 units and up 4 units is:
[tex](4,-6)\rightarrow(4+5,-6+4)=(9,-2)[/tex]The image point is (9, -2)
Suppose that the scores on a statewide standardized test are normally distributed with a mean of 63 and a standard deviation of 2. Estimate the percentage of scores that were(a) between 59 and 67. %(b) above 69. %(c) below 59. %(d) between 57 and 67. %
a) Percentage of scores that were between 59 and 67 = 95.45%
b) Percentage of scores above 69 = 0.135%
c) Percentage of scores below 59 = 2.2755%
d) Percentage of scores between 57 and 67 = 97.59%
Explanations:The mean, μ = 63
Standard deviation, σ = 2
When x = 59
[tex]\begin{gathered} z\text{ = }\frac{x-\mu}{\sigma} \\ z\text{ = }\frac{59-63}{2} \\ z\text{ = }-2 \end{gathered}[/tex]When x = 67
[tex]\begin{gathered} z\text{ = }\frac{x-\mu}{\sigma} \\ z\text{ = }\frac{67-63}{2} \\ \text{z = 2} \end{gathered}[/tex]P(59 < x < 67) = P(-2 < x < 2) = 0.9545
Probability that scores fall between 59 and 67 = 0.9545
Percentage of scores that were between 59 and 67 = 95.45%
b) above 69
P(x > 69)
[tex]\begin{gathered} z\text{ = }\frac{x-\mu}{\sigma} \\ z\text{ = }\frac{69-63}{2} \\ z\text{ = 3} \end{gathered}[/tex]P(x > 69) = P(z > 3) = 0.0013499
Percentage of scores above 69 = 0.135%
c) below 59
P(x < 59)
[tex]\begin{gathered} z\text{ = }\frac{59-63}{2} \\ z\text{ = -2} \end{gathered}[/tex]P(x < 59) = P(z < -2) = 0.02275
Percentage of scores below 59 = 2.2755%
d) between 57 and 67.
when x = 57
[tex]\begin{gathered} z\text{ = }\frac{57-63}{2} \\ z\text{ = -3} \end{gathered}[/tex]P(57 < x < 67) = P(-3 < x < 2) = 0.9759
Percentage of scores between 57 and 67 = 97.59%
which of the following is not arithmetic sequence? question in photo
Option C is correct.
Explanation:
A sequence is an arithmetic sequence when the difference between one term and the next term is the same. The first 10 counting numbers have a difference equal to 1. Skip counting by 5 has a difference equal to 5. Odd numbers have a difference equal to 2. But the prime numbers do not have the same difference. Thus, option c is not an arithmetic sequence.