Use the table. What percentage of the people surveyed were teachers who wanted a later start time?
Answers
The Solution.
The percentage of the people survey that were teachers that voted yes to start later is
[tex]\text{ }\frac{\text{ number of teachers that voted YES}}{\text{ Total number of people surveyed}}\times100[/tex]Which is
[tex]\frac{20}{75}\times100=0.266667\times100=26.6667\approx26.67\text{ \%}[/tex]b. The percentage of the people surveyed that were teachers is
[tex]\frac{\text{ number of teachers surveyed}}{\text{ Total number of people surveyed}}\times100[/tex]Which is
[tex]\frac{30}{75}\times100=0.4\times100=40\text{ \%}[/tex]Hence, the correct answer are:
a. 26.67% b. 40%
Related Questions
Identify the arc length of MA in terms of pi and rounded to the nearest hundredth.
Answers
To answer this question we will use the following formula for the arc length of a central angle θ degrees:
[tex]\begin{gathered} \frac{\theta}{180}\cdot\pi r, \\ \text{where r is the circumference's radius.} \end{gathered}[/tex]Assuming that Y is the circumference's center we get:
[tex]m\hat{AM}+m\hat{MH}=180^{\circ}.[/tex]Substituting mMH=88degrees we get:
[tex]m\hat{AM}+88^{\circ}=180^{\circ}\text{.}[/tex]Therefore:
[tex]\text{m}\hat{\text{AM}}=92^{\circ}\text{.}[/tex]Then the arc length of MA is:
[tex]\frac{92}{180}\cdot\pi\cdot16m\approx8.18\pi m\approx25.69m\text{.}[/tex]Answer: First option.
Write an explanation of how you solved the problem. Write the explanation so that another student could follow your thought process.
Answers
Solution
- The solution steps are given below:
[tex]\begin{gathered} \sqrt{125} \\ 125\text{ can be written as:} \\ 125=5\times25 \\ \\ \text{ Thus, we have:} \\ \sqrt{125}=\sqrt{5\times25} \\ \\ \text{ Using the surd rule that:} \\ \sqrt{ab}=\sqrt{a}\times\sqrt{b} \\ \text{ We have:} \\ \\ \sqrt{5\times25}=\sqrt{5}\times\sqrt{25} \\ \\ \therefore\sqrt{125}=\sqrt{5}\times5 \\ \\ =5\sqrt{5} \end{gathered}[/tex]Final Answer
The answer is
[tex]5\sqrt{5}[/tex]Convert this rational number
to its decimal form and round
to the nearest tenth
2/5
Answers
Answer:
0.4
Step-by-step explanation:
First, divide 2 by 5. You should but a decimal because 5 doesn't go into 2, so you get 20. 5 does go into 20, 4 times.
Use substitution to determine the solution of the system of equations. Write the solution as an ordered pair.x + 2y = 14y = 3x – 14solution =
Answers
the first step:
we will rewrite the second equation to look like the first one
y=3x-14
lets subtract 3x from both sides
y-3x=-14
-3x+y=-14
our system of equations will look like
x+2y=14
-3x+y=-14
however, for a substitution method we can use:
x+2(3x-14)=14
i multiply 2 with the parenthesis
x+6x-28=14
7x=28+14
7x=42, x=42/7=6
x=6
y=3x-14
y=(3x6)-14=18-14=4
solution= 6,4
A zero-coupon bond is a bond that is sold now at a discount and will pay its face value at the time when it matures; no interest payments are made.A zero-coupon bond can be redeemed in 20 years for $10,000. How much should you be willing to pay for it now if you want the following returns?(a) 8% compounded daily(b) 8% compounded continuously
Answers
EXPLANATION:
We are given a zero-coupon bond that will be worth $10,000 if redeemed in 20 years time at an annual rate of 8% compounded;
(a) Daily
(b) Continuously
The formula for compounding annually is given as follows;
[tex]A=P(1+r)^t[/tex]Here the variables are;
[tex]\begin{gathered} P=initial\text{ investment} \\ A=Amount\text{ after the period given} \\ r=rate\text{ of interest} \\ t=time\text{ period \lparen in years\rparen} \end{gathered}[/tex]Note that this zero-coupon bond will yield an amount of $10,000 after 20 years at the rate of 8%. This means we already have;
[tex]\begin{gathered} A=10,000 \\ r=0.08 \\ t=20 \end{gathered}[/tex](a) For interest compounded daily, we would use the adjusted formula which is;
[tex]A=P(1+\frac{r}{365})^{t\times365}[/tex]This assumes that there are 365 days in a year.
We now have;
[tex]10000=P(1+\frac{0.08}{365})^{20\times365}[/tex][tex]10000=P(1.00021917808)^{7300}[/tex][tex]10000=P(4.95216415047)[/tex]Now we divide both sides by 4.95216415047;
[tex]P=\frac{10000}{4.95216415047}[/tex][tex]P=2019.31916959[/tex]We can round this to 2 decimal places and we'll have;
[tex]P=2019.32[/tex](b) For interest compounded continuously, we would use the special formula which is;
[tex]A=Pe^{rt}[/tex]Note that the variable e is a mathematical constant whose value is approximately;
[tex]e=2.7183\text{ \lparen to }4\text{ }decimal\text{ }places)[/tex][tex]10000=Pe^{0.08\times365}[/tex][tex]10000=Pe^{29.2}[/tex]With the use of a calculator we have the following value;
[tex]\frac{10000}{e^{29.2}}=P[/tex]Write down the first five terms of the sequence an=(n+4)!2n2+6n+7a1 = a2 = a3 = a4 = a5 =
Answers
Step 1
Given;
[tex]\begin{gathered} a_n=\frac{(n+4)!}{2n^2+6n+7} \\ n=1,2,3,4,5 \end{gathered}[/tex]Step 2
[tex]a_1=\frac{(1+4)!}{2(1)^2+6(1)+7}=\frac{120}{15}=8[/tex][tex]a_2=\frac{(2+4)!}{2(2)^2+6(2)+7}=\frac{720}{27}=\frac{80}{3}[/tex][tex]a_3=\frac{(3+4)!}{2(3)^2+6(3)+7}=\frac{5040}{43}[/tex][tex]a_4=\frac{(4+4)!}{2(4)^2+6(4)+7}=\frac{8!}{63}=640[/tex][tex]a_5=\frac{(5+4)!}{2(5)^2+6(5)+7}=\frac{9!}{87}=\frac{120960}{29}[/tex]coefficient. | The (blank) factor that appears before a varible. constant of proportionality. | The value of the ratio between (blank) quantities that are in a proportional relationship. equation.| A mathematical statement equating two (blank). proportional. | a relationship between two quantities such that as one value increases or (blank), the other value will increase or decrease by the same multiple.
Answers
1st statement = constant number factor
2nd statement =variable
3rd statement = expression
4 th statement = decreses
About 53.6% of the students in the class got a AA. If 15 students got an A, estimate the number of students in the class B. If the students got an A, find the actual number of students in the class
Answers
For A.
we have that 15 students got an A that represents the 53.6%
In this case w need to find the 100%
15 is 53.6%
x is 100%
x is the total number of students in the class
[tex]x=\frac{\frac{100}{100}\times15}{\frac{53.6}{100}}[/tex][tex]x=\frac{15\cdot100}{53.6}[/tex][tex]x=27.98[/tex]A woman is traveling in her car at 50 miles per hour. How far will that woman travel in fifteen minutes if her speed is constant? (Distance = Rate x Time)
Answers
we know that
Distance=ratex Time
where
Rate=50 mph
Time=15 min
Convert the time to hours
1 hour=60 min
so
15 min=15/60=0.25 hours
substitute in the formula
Distance=50x0.25
Distance=12.50 milesHe starts by finding the sum of the exterior angles of a Pentagon, which isand then he solves to find that x is
Answers
The sum of the exterior angle of all polygons is same.
This is equal to 360
So, when we add all the exterior angles, this is equal to 360
Thus, mathematically;
[tex]\begin{gathered} x\text{ +2x + 78 + 77 + 3x+3 = 360} \\ \\ 6x\text{ + 158 = 360} \\ \\ 6x\text{ = 360-158} \\ \\ 6x\text{ = 202} \\ \\ x\text{ = }\frac{202}{6} \\ \\ x\text{ = 33}\frac{2}{3} \end{gathered}[/tex]The florist charge $31.75 for eight roses and five carnations. For one rose and three carnations,it costs $5.75. What is the cost for each type of flower?
Answers
Let's use the variable x to represent the cost of one rose and y to represent the cost of one carnation.
If 8 roses and 5 carnations cost $31.75, we can write the following equation:
[tex]8x+5y=31.75[/tex]If 1 rose and 3 carnations cost $5.75, we can write a second equation:
[tex]x+3y=5.75[/tex]From the second equation, we have x = 5.75 - 3y.
Using this value of x in the first equation, we have:
[tex]\begin{gathered} 8(5.75-3y)+5y=31.75\\ \\ 46-24y+5y=31.75\\ \\ -19y=31.75-46\\ \\ -19y=-14.25\\ \\ y=0.75 \end{gathered}[/tex]Now, calculating the cost of one rose, we have:
[tex]x=5.75-3y=5.75-3\cdot0.75=5.75-2.25=3.5[/tex]Therefore the cost of one rose is $3.50 and the cost of one carnation is $0.75.
Please help me thank you I’ll send tutor a chart that goes with the question
Answers
Answer:
As time increases, the number of flyers Justin has increased.
3 flyers per minute
62 flyers at t
Explanation:
Based on the table, we can see that when the time is greater the number of flyers is also greater. So, as time increases, the number of flyers Justin has increased.
Then, to know the rate, we need to use two columns of the table, so the rate will be equal to:
[tex]\frac{Flyers_2-Flyers_1}{Time_2-Time_1}[/tex]So, if we replace Time1 by 10, Flyers1 by 92, Time2 by 15, and Flyers2 by 107, we get:
[tex]\text{rate}=\frac{107-92}{15-10}=\frac{15}{5}=3[/tex]Therefore, the rate is 3 flyers per minute.
Now, to know how many flyers he has when he started printing, we need to extend the table. We see that every 5 minutes, the flyers increase by 15. So, at 0 and 5 minutes the number of flyers was:
Time 0 5 10
Flyers 77-15 = 62 92-15 = 77 92
Therefore, the number of flyers that he has when he started printing was the number of flyers at 0 minutes. So the answer of part b is 62 flyers
Deion measure the volume of a sink basin by modeling it as a hemisphere. Deion measures its diameter to be 28 3/4 inches. Find the sink’s volume in cubic inches. Round your answer to the nearest tenth if necessary.
Answers
28.75We are asked to determine the volume of a hemisphere of diameter 28 3/4 in.
A hemisphere is half a sphere therefore, its volume is half the volume of a sphere:
[tex]V=\frac{1}{2}(\frac{4\pi r^3}{3})=\frac{2\pi r^3}{3}[/tex]Where "r" is the radius. Since the radius is half the diameter we have that:
[tex]r=\frac{D}{2}=\frac{28\frac{3}{4}}{2}[/tex]We will convert the mixed fraction into a standard fraction using the following:
[tex]28\frac{3}{4}=28+\frac{3}{4}=28.75[/tex]Substituting in the formula for the radius:
[tex]r=\frac{28.75in}{2}=14.38in[/tex]Now, we substitute the value of the radius in the formula for the volume:
[tex]V=\frac{2\pi(14.38in)^3}{3}[/tex]Solving the operations:
[tex]V=6221.3in^3[/tex]Therefore, the volume is 6221.3 cubic inches.
What does the constant 0.98 reveal about the rate of change of the quantity?
Answers
The given function is:
[tex]f(x)=780(0.98)^{10t}[/tex]It is required to state what the constant 0.98 reveal about the rate of change of the quantity by completing the given statement:
The function is exponentially at a rate of every .
Notice from the given exponential function that the factor is 0.98.
Since it is less than 1, it implies that it is an exponential DECAY.
Subtract the factor 0.98 from 1 to get the decay rate:
[tex]1-0.98=0.02=0.02\times100\%=2\%[/tex]Unit of t: decades
The exponent is 10t.
The reciprocal is 1/10.
Hence, the time frame of rate is: 1/10 of a decade.
This is equivalent to a year.
The complete statement is, therefore:
The function is decaying exponentially at a rate of 2% every year.
I did 4/cos(62°) but it didn't give me any of the answer options
Answers
we have that
tan(62)=x/4 ------> by opposite side divided by the adjacent side
solve for x
x=4*tan(62)
x=7.5 unitsCan you help me please?A. How can Marc provide proof that his mighty shot actually hung in the air for 15 seconds? Or is this just another one of his lies?B. How long did the ball actually hang in the air?
Answers
The given formula for Marc's shot is:
[tex]h(x)=-16x^2+200x[/tex]a. To prove that the shot actually hung in the air for 15 seconds, we need to replace x=15 in the formula and solve for h, as follows:
[tex]\begin{gathered} h(15)=-16(15)^2+200(15) \\ h(15)=-16\times225+3000 \\ h(15)=-3600+3000 \\ h(15)=-600 \end{gathered}[/tex]As the height is negative, it means after 15 seconds the ball already hit the ground, because the ground is located at h=0. Then this result proves that this is just another one of Marc's lies.
b. To find how long the ball actually hung in the air, we need to find the x-values that makes h=0, as follows:
[tex]0=-16x^2+200x[/tex]We have a polynomial in the form: ax^2+bx+c=0, where a=-16, b=200 and c=0.
We can use the quadratic formula to solve for x:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ x=\frac{-200\pm\sqrt[]{(200)^2-4(-16)(0)}}{2(-16)} \\ x=\frac{-200\pm\sqrt[]{40000+0}}{-32} \\ x=\frac{-200\pm\sqrt[]{40000}}{-32} \\ x=\frac{-200\pm200}{-32} \\ x=\frac{-200+200}{-32}=\frac{0}{-32}=0\text{ and }x=\frac{-200-200}{-32}=\frac{-400}{-32}=12.5 \end{gathered}[/tex]Then the two x-values are x=0 and x=12.5.
The starting time is 0 and the end time when the ball hit the ground is x=12.5.
The ball actually hung in the air 12.5 seconds.
19. Does the following system have a unique solution? Why?4x-6y=-7-2x + 3y = 18
Answers
Given,
The equations are
[tex]\begin{gathered} 4x-6y=-7..............(1) \\ -2x+3y=18............(2) \end{gathered}[/tex]To find: Does the following system have a unique solution? Why?
Solution:
The determinant of the given equations are
[tex]\begin{gathered} \begin{bmatrix}{4} & {-6} \\ {-2} & {3}\end{bmatrix}=\begin{bmatrix}{-7} & {} \\ {18} & {}\end{bmatrix} \\ 4\times3-(-6\times-2) \\ =12-12 \\ =0 \end{gathered}[/tex]For unique solutions, the condition is
[tex]\begin{gathered} \frac{a_1}{a_2}=\frac{b_1}{b_2} \\ \frac{4}{-2}=\frac{-6}{3} \\ -2=-2 \end{gathered}[/tex]Condition satisfied.
Hence, the given equations have a unique solution.
what type of angle is angle 98 degrees
Answers
Answer:
Obtuse
Step-by-step explanation:
Since it is larger than 90 degrees but less than 180 degrees, an angle of 98 degrees is considered to be obtuse.
Hope this helps! :)
An equation is shownn-5=17What is the value for n that makes the equation true
Answers
Simplify the equation n-5=17.
[tex]\begin{gathered} n-5=17 \\ n=17+5 \\ =22 \end{gathered}[/tex]So value of n is 22.
6 a. Sketch a reflection triangleAABC about the line y - X.Label the image AA'B'C'. I know youdon't have graph paper, just sketch.b. What are the coordinates of C'?
Answers
Notice that we are asked to find the image of a triangle that
The following list gives the number of pets for each of 14 students.1, 4, 4, 2, 3, 4, 3,0,0,0,3,0,0,1Send data to calculatorFind the modes of this data set.If there is more than one mode, write them separated by commas.If there is no mode, click on "No mode."
Answers
Given:
The number of pets for each of 14 students is,
1, 4, 4, 2, 3, 4, 3,0,0,0,3,0,0,1.
Mode: It is the number that appears the most.
In the given data,
0 appears the most that is for 5 times.
So, the mode of the given data is 0.
Answer: Mode is 0.
ΔABC - ΔDEE B D X = [?] Ente
Answers
Triangle ABC = Triangle DEF
Both triangles are similar
Hence, we will be using the similarity theorem
AB / DE = BC / EF
x / 3 = 8/6
Cross multiply
6 * x = 8 * 3
6x = 24
Divide both sides by 6
6x / 6 = 24/ 6
x = 24/ 6
x = 4
The answer is 4
A triangle has side lengths of 13 mm 18 mm in 14 mm classify it as a cute obtuse or right
Answers
Given
Sides
13 mm, 18 mm, 14 mm
Using the Pythagorean theorem, compare the square of the longest side to the sum of the other two sides
[tex]\begin{gathered} a^2+b^2\text{ ? }c^2 \\ 13^2+14^2\text{ ? }18^2 \\ 169+196\text{ ? }324 \\ 365>324 \end{gathered}[/tex]Since the sum of the square of two sides is greater than the square of the longest side, the triangle is acute.
Hi! I just wanted to know may you please help me on knowing how to solve an inequality?
Answers
When given an inequality, we solve it as if we were solving an equation but there is one important consideration: If we divide or multiply by a negative number the inequality gets inverted.
Recall that to solve an equation we add, subtract, multiply, etc. to isolate the variable, in the case of inequalities we do the same process.
Example 1. x+58>36, to solve the inequality for x, we have to put the variable by itself on one side of the inequality, to do that we have to eliminate whatever is next to the variable, in the case 58 is adding, therefore, we must subtract it, but to not alter the inequality we do the same to both sides of the inequality.
Adding -58 to both sides of the inequality we get:
[tex]\begin{gathered} x+58-58>36-58, \\ x>-22. \end{gathered}[/tex]Therefore, the solution to this inequality is x>-22.
Example 2. -3x+3<4:
Adding -3 to both sides of the inequality we get:
[tex]\begin{gathered} -3x+3-3<4-3, \\ -3x<1. \end{gathered}[/tex]Dividing by -3 we get:
[tex]\begin{gathered} \frac{-3x}{-3}>\frac{1}{-3}, \\ x>-\frac{1}{3}\text{.} \end{gathered}[/tex]Use sigma notation to represent the sum of the first eight terms of the following sequence: 4,7. 10,
Answers
To answer this question, we need to check the kind of sequence here. We have that the first three elements of the sequence are:
[tex]4,7,10[/tex]If we have the difference between the second element and the first element, and the difference between the third element and the second element, we have:
[tex]7-4=3,10-7=3[/tex]Thus, we have a common difference of 3. This is an arithmetic sequence. Then, we also have that:
[tex]a_n=a_1+(n-1)d[/tex]This is the formula for finding a general term in an arithmetic sequence, where:
• an is any term, n, in the sequence.
,• a1 is the first term in the sequence.
,• n the number of the term in the sequence.
,• d is the common difference of the sequence.
In this way, we have:
a1 = 4
d = 3
What should be done to these equations to solve the system of equations by elimination? [tex]2x + y = 12 \\ 5x - 3y = -3[/tex]A: Divide the first equation by 2.B: Divide the second equation by 2.C: Multiply the first equation by 3.D: Multiply the second equation by 3.
Answers
ANSWER:
C: Multiply the first equation by 3.
STEP-BY-STEP EXPLANATION:
We have the following the system of equations:
[tex]\begin{gathered} 2x+y=12 \\ 5x-3y=-3 \end{gathered}[/tex]To solve a system of equations by means of the elimination method, it must coincide in such a way that when adding equation 1 and equation 2, one of the variables are eliminated, that is, they are canceled.
Therefore, the option in this case is to multiply the first equation by 3, in this way, the first equation remains as 3y and the second remains as -3y, since when adding them, it is 0 and the variable y is canceled.
The figure shows the layout of a symmetrical pool in a water park. What is the area of this pool rounded to the tens place? Use the value = 3.14.
Answers
EXPLANATION:
Given;
We are given a symmetrical pool as indicated in the attached picture.
The pool consists of two sectors and two triangles and each pair has the same dimensions.
The dimensions are as follows;
[tex]\begin{gathered} Sector: \\ Radius=30 \\ Central\text{ }angle=2.21\text{ }radians \end{gathered}[/tex][tex]\begin{gathered} Triangle: \\ Slant\text{ }height=30 \\ Vertical\text{ }height=25 \\ Base=20 \end{gathered}[/tex]Required;
We are required to calculate the area of the pool.
Step-by-step solution;
We shall begin by calculating the area of the sector and the formula for the area of a sector is;
[tex]\begin{gathered} Area\text{ }of\text{ }a\text{ }sector: \\ Area=\frac{\theta}{2\pi}\times\pi r^2 \end{gathered}[/tex]Where the variables are;
[tex]\begin{gathered} \theta=2.21\text{ }radians \\ r=30 \\ \pi=3.14 \end{gathered}[/tex]We now substitute and we have the following;
[tex]Area=\frac{2.21}{2\pi}\times\pi\times30^2[/tex][tex]Area=\frac{2.21}{2}\times900[/tex][tex]Area=994.5ft^2[/tex]Since there are two sectors of the same dimensions, the area of both sectors therefore would be;
[tex]Area\text{ }of\text{ }sectors=994.5\times2[/tex][tex]Area\text{ }of\text{ }sectors=1989ft^2[/tex]Next we shall calculate the area of the triangles.
Note the formula for calculating the area of a triangle;
[tex]\begin{gathered} Area\text{ }of\text{ }a\text{ }triangle: \\ Area=\frac{1}{2}bh \end{gathered}[/tex]Note the variables are;
[tex]\begin{gathered} b=20 \\ h=25 \end{gathered}[/tex]The area therefore is;
[tex]Area=\frac{1}{2}\times20\times25[/tex][tex]Area=\frac{20\times25}{2}[/tex][tex]Area=250[/tex]For two triangles the area would now be;
[tex]Area\text{ }of\text{ }triangles=250\times2[/tex][tex]Area\text{ }of\text{ }triangles\text{ }equals=500ft^2[/tex]Therefore, the area of the pool would be;
[tex]\begin{gathered} Area\text{ }of\text{ }pool: \\ Area=sectors+triangles \end{gathered}[/tex][tex]\begin{gathered} Area=1989+500 \\ Area=2489ft^2 \end{gathered}[/tex]Rounded to the tens place, we would now have,
ANSWER:
[tex]Area=2,490ft^2[/tex]Option D is the correct answer
Create a list of steps, in order, that will solve the following equation.5(x-3)² + 4 = 129Solution steps:
Answers
We want to solve the equation
[tex]5\cdot(x-3)^2+4=129[/tex]To solve this equation, we must isolate the variable on one side of the equation. So, noticed that the term with the x has multiple things (it is raised to the second power, multiplied by five, etc) so we begin by subtracting 4 on both sides. (Step 1)
So we get
[tex]5\cdot(x-3)^2=129\text{ -4 =125}[/tex]Now we divide both sides by 5 (Step 2), so we get
[tex](x-3)^2=\frac{125}{5}=25[/tex]now we take the square root on both sides (Step 3). Then we get
[tex]\sqrt[]{(x-3)^2}=\sqrt[]{25}=5=x\text{ -3}[/tex]Finally, we add 3 on both sides (Step 4). Then we get
[tex]x=5+3=8[/tex]Six less than x equals twenty-two.
Answers
What do I need to do?
Write the word expression in algebraic expression?
16.- x - 6 = 22
x = 22 + 6
x = 28
17.-
y/4 = -10
y = 4(-10)
y = -40
19.- h - 7 = -9
h = -9 + 7
h = -2
20.- x + y = -7
if x = 11
11 + y = -7
y = -7 - 11
y = -18
Charlie is saving money to buy a game. So far he has saved $16, which is one-half of the total cost of the game. How much does the game cost?
Answers
Answer:
If $16 is just one-half of the total cost of the game, then the cost of the game must be $32.
Explanation:
See the illustration below for 1/2.
Charlie has saved $16 already which is 1/2 of the total game cost.
As we can see above, Charlie needs 1/2 more to be able to buy a game. This means Charlie needs to save an additional $16.
In total, Charlie must saved $16 + $16 = $32 to buy the game.
The game cost must be $32.