Calculate the relative frequency of the second class determine the answer as a simplified fraction
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Given: 1105,1180,1056,1102,1295,1313,1146,1248,1133,1059,1352,1317,1111,1025,1125,1270,1362,1103,1357,1062,1331
Find: relative frequency of second class.
Explanation: the second class varies from 1072-1131
1105,1102,1111,1125,1103. are the numbers lies from 1072-1131
these are 5 number so their frequency is 5.
and total numbers are 21 . so the relative frequency is
[tex]\frac{5}{21}=0.23[/tex]Final answer: the relative frequency of second class is 0.23
Related Questions
Troy opened a savings account and deposited 600.00 as principal the account earns 11%interest compounded monthly what is the balance after 3 years
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The height of a houseplant is normally distributed with a mean of 12 inches and a standard deviation of 2 inches. Using the empirical rule, what is the probability that a houseplant will be longer than 16 inches? a0.975 b0.95 c0.05 d0.025
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Solution:
From the information given:
[tex]\begin{gathered} mean,\text{ }\mu=12\text{ inches} \\ Standard\text{ deviation, }\sigma=2\text{ inches} \\ sample,\text{ x = 16} \end{gathered}[/tex]Calculate the z-score
[tex]z=\frac{x-\mu}{\sigma}=\frac{16-12}{2}=\frac{4}{2}=2[/tex]The empirical state as shown below
We can see in the illustration above that population within 2 standard deviation is 95% (i.e 0.95).
However, The question says to find the probability that a houseplant will be longer than 16 inches, that is above 2 standard deviation from our calculation
This gives 1-0.95 = 0.05
Could you please help me with this problem? It’s a bit blurry. From a computer screen.
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First find out the selling price of the clarinet, this is:
Discounted price / ( 1 - discount rate)
[tex]\frac{93.60}{(1-20\%)}=\frac{93.60}{(1-0.20)}=\frac{93.60}{0.80}=117[/tex]This selling price is 30% more than what the store paid so the price the store paid was:
[tex]\frac{117}{1+0.30}=\frac{117}{1.30}=90[/tex]Answer: the store pay for the clarinet $90
please help What is the First Step in solving this equation (isolating de variable)?=10
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The first step in solving an equation (isolating the variable ) = 10
First, Is to do common thing to the both sides either by adding , substracting or dividing
Example: 2a = 10
To isolate variable a
Divivde the both sides by the coefficient of a which is 2
2a / 2 = 10/2
a = 5
I need help solving this problem trigonometric function don't kind 22.)
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Draw a diagram to visualize the situation:
Recall the definition of the sine of an angle:
[tex]\sin \theta=\frac{\text{Side opposite to }\theta}{\text{ Hypotenuse}}[/tex]In this case, the side opposite to θ has a measure of 80m, while the hypotenuse has a measure of 150m. Then:
[tex]\begin{gathered} \sin \theta=\frac{80m}{150m} \\ \Rightarrow\sin \theta=\frac{8}{15} \end{gathered}[/tex]Use the inverse sine function to isolate θ, and use a calculator to find its value:
[tex]\Rightarrow\theta=\sin ^{-1}(\frac{8}{15})=32.23095264\ldots[/tex]Therefore, up to two decimal places:
[tex]\theta=32.23^{\circ}[/tex]In the figure below ABC ~ YZX.Fund tanY, cosY, and sinY.Round your answers to the nearest hundredth.I’m super confused so please help.
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We know that the given triangles are similar. That means Angles Y and B are congruent. So, our job is to find Angle B first using trigonometric reasons. Let's use the tangent which is equal to the ratio between the opposite leg and the adjacent leg to angle B
[tex]\begin{gathered} \text{tan(B)=}\frac{14.7}{14}=1.05=\tan Y \\ B=\tan ^{-1}(1.05)\approx46.4 \end{gathered}[/tex]Which means angle Y is also 46.4°, approximately. Also, the tangent of Y is 1.05 too, because Y and B are congruent. Now, let's find the sine and cosine of B or Y (remember they are equal).
[tex]\sin B=\frac{14.7}{20.3}\approx0.72=\sin Y[/tex][tex]\cos B=\frac{14}{20.3}\approx0.69=\cos Y[/tex]Therefore, the tanY is 1.05, sinY is 0.72 and cosY 0.69.|Pythagorean TheoremFor the following right triangle, find the side length x. Round your answer to the nearest hundredth.810
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A right triangle satisfies the Pythagorean Theorem:
[tex]c^2=a^2+b^2[/tex]where c is the hypotenuse (larger side), while a and b are the legs (smaller sides).
In our case:
• c = x
,• a = 8
,• b = 10
Replacing the values we get:
[tex]x^2=8^2+10^2[/tex]Solving for x:
[tex]\sqrt[]{x^2}=\sqrt[]{8^2+10^2}[/tex][tex]x^{}=\sqrt[]{64+100}[/tex][tex]x=\sqrt[]{164}\approx12.81[/tex]Answer: 12.81
Consider functions fand gUsing a table of values, what is the approximate solution to the equation f(I) = g(I| to the nearest quarter of a unit?
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Given:
Two functions f(x) and g(x) are given.
[tex]\begin{gathered} f(x)=\frac{x-1}{x^2+x-1} \\ g(x)=3^x-2 \end{gathered}[/tex]Required:
Find the approximate solution to the equation f(x)=g(x) to the nearest quarter of the unit.
Explanation:
(a) substitute x= 0.50in f(x) and g(x).
[tex]\begin{gathered} f(x)=\frac{0.5-1}{(0.5)^2+0.5-1} \\ f(x)=\frac{-0.5}{-0.25} \\ f(x)=2 \end{gathered}[/tex][tex]\begin{gathered} g(x)=3^{0.5}-2 \\ g(x)=-0.268 \end{gathered}[/tex](b) substitute x=-1.75 in f(x) and g(x).
[tex]\begin{gathered} f(x)=\frac{-1.75-1}{(-1.75)^2-1.75-1} \\ f(x)=\frac{-2.75}{0.38125} \\ f(x)=-7.2 \end{gathered}[/tex][tex]\begin{gathered} g(x)=3^{-1.75}-2 \\ g(x)=-1.854 \end{gathered}[/tex](c) substitute x= 0.75 in f(x) and g(x).
[tex]\begin{gathered} f(x)=\frac{0.75-1}{(0.75)^2+0.75-1} \\ f(x)=\frac{-0.25}{0.3125} \\ f(x)=-0.8 \end{gathered}[/tex][tex]\begin{gathered} g(x)=3^{0.75}-2 \\ g(x)=2.795 \end{gathered}[/tex](d) substitute x= -2.25 in f(x) and g(x).
[tex]\begin{gathered} f(x)=\frac{-2.25-1}{(-2.25)^2-2.25-1} \\ f(x)=\frac{-3.25}{1.8125} \\ f(x)=-1.793 \end{gathered}[/tex][tex]\begin{gathered} g(x)=3^{-2.25}-2 \\ g(x)=-1.916 \end{gathered}[/tex]we can observe that at x = -2.25 the functions f(x) and g(x) has the approximate values.
Final Answ
1. What are the coordinates of the vertices of triangle J"K"L"? 2. Write ONE algebraic rule that represents triangle JKL being translated to triangle J"K"L".
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The triangle JKL has vertices J(-3,-3), K(-4,2), and L(2,1)
1) This triangle was first translated using the rule:
(x,y) → (x+2,y-3)
This means it was translated 2 units to the right and 3 units down.
JKL → J'K'L'
J(-3,-3) → J'(-3+2,-3-3) = (-1,-6)
K(-4,2) → K'(-4+2,2-3) = (-2,-1)
L(2,1) → L'(2+2,1-3) = (4,-2)
The resulting triangle J'K'L' has vertices J'(-1,-6), K'(-2,-1), and L'(4,-2)
2) The triangle J'K'L' was translated using the algebraic rule
(x,y) → (x-2,y-1)
This indicates it was translated 2 units to the left and one unit down.
The translation is
J'K'L' → J''K''L''
J'(-1,-6) → J''(-1-2,-6-1) = (-3,-7)
K'(-2,-1) → K''(-2-2,-1-1) = (-4,-2)
L'(4,-2) → L''(4-2,-2-1) = (2,-3)
The triangle J''K''L'' has vertices J''(-3,-7), K''(-4,-2), and L''(2,-3)
To determine one algebraic rule that represents the translation from JKL to J''K''L'' directly, you have to compare the coordinates of the vertices of both triangles:
JKL → J''K''L''
J(-3,-3) → J''(-3,-7)
K(-4,2) → K''(-4,-2)
L(2,1) → L''(2,-3)
You have to compare the x-coordinates and the y-coordinates of both triangles.
If you compare the x-coordinates you will notice that they are the same, this means that to make a direct translation from JKL to J''K''L'' you don't have to make any horizontal movements.
If you compare the y-coordinates you will see that they are different, so between both triangles, there was a vertical movement done. To determine the length of said movement, you have to calculate the difference between the y-coordinate of one vertex of J''K''L'' and the corresponding y-coordinate of the vertex of JKL.
For example compare J(-3,-3) and J''(-3,-7)
[tex]y_{J^{\doubleprime}}-y_J=-7-(-3)=-4[/tex]→ So we can conclude that to move JKL to J''K''L'' you have to make a vertical translation 4 units down. The algebraic rule that represents this translation is
(x,y) → (x,y-4)
The figure is not drawn to scale. ABCD is a rectangle and AC and BD are diagonals. The ratio of
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If we take the measures of the angles p and q as x, we will notice that the angles m and n correspond to 5x and since it is a straight line, it must be 5x = 180 °
solving for x we obtain 36 °
and since the measures of p q and the angle we are looking for are a rectangle, they must add 90 °
To find the unknown measure we can subtract the measures of the angles p and q from 90 °, this is 90-36-36
The measure of the EBF angle must be 18°.
Evaluate. (Be sure to check by differentiating!) 8 6x dx (3+x0)* | (3+x0)86x® dx =0
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five minus what equals 11
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Let's call that what w.
We get that:
[tex]5-w=11[/tex]Solving for w,
[tex]5-11=w\rightarrow w=-6[/tex]Thereby, 5-(-6)=11
For the diagram below, if < 6 =141 °, what would be the measure of < 4 ?
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From the information given,
angle 6 = 141
angle 5 = 180 - angle 6(Because they are on a straight line)
angle 5 = 180 - 141 = 39
angle 5 = angle 3(Alternate angles are congruent)
ang;e 3 = 39
angle 4 = 180 - 39(Because they are on a straight line)
angle 4 = 141 degrees
Identify the number of solutions of the system of linear equations.x + 3y - z=2x+y-z=03x + 2y - 3z = -1• no solution• exactly one solution•infinitely many solutionsSolve the system using the elimination method. If there are infinitely many solutions, write the ordered triple in terms of z. If there is nosolution, leave each box empty.
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Given the equations:
[tex]\begin{gathered} x+3y-z=2\ldots\ldots.\ldots\ldots....\ldots\ldots\ldots\ldots\ldots\text{.}(1) \\ x+y-z=0\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots..\ldots..\ldots(2) \\ 3x+2y-3z=-1\ldots.\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots(3) \end{gathered}[/tex]We want to solve them by elimination.
[tex]undefined[/tex]what does a quadratic formal look like
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The quadratic formula is used to find the solution of quadratic functions.
Given any quadratic function in standard form:
[tex]ax^2+bx+c=0[/tex]Where
a is the coefficient of the quadratic term
b is the coefficient of the x-term
c is the constant of the quadratic function
The quadratic formula has the following shape:
[tex]x=\frac{b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Find the sum. You may find using a number line to be helpful Express your answer as a simplified mixednumber, if necessary3+1 1/2
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To perform this sum, we will use a drawing of the real line
Now, we are going to add the number 1 1/2. The number 1 1/2 means that we have 1 and a half units. So, adding up one and a half units is equivalent to first adding up one unit and then the other half. So, if we add first one unit, we get
three pears weigh the same as an orange. an orange weighs as much as eighteen raspberry.How many raspberries weigh the same as a pear?
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Answer:
6 raspberries.
Step-by-step explanation:
If one orange is 18 raspberries, and three pears weighs one orange, you need to divide 3 by 18.
3 pears = 1 orange
1 orange = 18 raspberries
1 pear = 6 raspberries
What is an equation of the line of symmetry for the parabola in the accompanying diagram?
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Given the figure of a parabola:
As shown:
The vertex of the parabola is the point ( 3, -1)
So, the line of symmetry will be the line: x = 3
The answer is option 3) x = 3
The dimensions of a track used for cycling competitions, called a velodrome, ismeasured in meters. Addison currently bikes in a 7.5 kilometer race, which is 30 lapsaround the track, or a total of approximately 4.65 miles. For the next race, Addisonwould like to bike in the longer race, which is 40 laps. Approximately how many miles isthis race?
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Given:
The laps are 30
Total distance = 7.5 Kilometer
Find-:
The distance when bikes laps are 40
Explanation-:
The radius of the bike = x
The distance of one lap
[tex]=2\pi r[/tex]Where
[tex]r=\text{ Radius}[/tex]The distance of 30 laps is:
[tex]\begin{gathered} =2\pi r\times(30) \\ \\ =60\pi r \end{gathered}[/tex]The distance is 75 kilometers.
[tex]60\pi r=7.5[/tex]A kilometer is 4.65 miles.
So, distance is:
[tex]\begin{gathered} =7.5\times4.65 \\ \\ =34.875\text{ miles} \end{gathered}[/tex]The radius is:
[tex]\begin{gathered} 60\pi r=34.875\text{ miles} \\ \\ r=\frac{34.875}{60\times\pi} \\ \\ r=\frac{0.58125}{\pi} \\ \\ r=0.185\text{ miles} \end{gathered}[/tex]The distance of 40 laps is:
[tex]\begin{gathered} \text{ Distance }=40\times2\pi r \\ \\ \text{ Distance }=80\pi r \\ \\ \text{ Distance }=80\times\pi\times0.185 \\ \\ \text{ Distance }=46.5\text{ miles} \end{gathered}[/tex]The distance of the race is 46.5 miles
Answer:
The distance of the race is 46.5 miles
Step-by-step explanation:
Determine which statement correctly compares the following numbers more than one choice may be correct 756.36 973.98
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Answers:
a. The 7 in the number labeled B is 1/10 the value of 7 in the number labeled A.
c. The 3 in the number labeled B is 10 times the value of 3 in the number labeled A.
Explanation:
Number A is 756.36, so each number has the following value:
7: 700
5: 50
6: 6
3: 0.3
6: 0.06
In the same way, number B is 973.98, so each number has the following value
9: 900
7: 70
3: 3
9: 0.9
8: 0.08
Then, for each option, we get:
a. Correct because number 7 in B has a value of 70 and 70 is 1/10 of 700, which is the value of 7 in A.
b. Incorrect, the value of 3 in A is 0.3 and the value in B is 3.
c. Correct, 3 is 10 times 0.3 because 0.3 x 10 = 3. Then, the value of 3 in B is 10 times the value of 3 in A.
d. Incorrect, because 3 is not 1/10 of 0.3
So, the answers are
a. The 7 in the number labeled B is 1/10 the value of 7 in the number labeled A.
c. The 3 in the number labeled B is 10 times the value of 3 in the number labeled A.
A town's population has been growing linearly. In 2003 the population was 54,000. The population has been growing by 2200 people each year. Write an equatich for the population, P, 2 years after 2003. P = Use the formula to find the population in 2009:
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Since the population grows by 2200 people each year, after x years it grows:
2200 * x = 2200x
Now, since it was already 54000 in 2003, we have to add this value to the growth in order to find the total population P, x years after the year 2003. Thus, it is given by:
P = 54000 + 2200x
Then, using this formula, we can use x = 6 to find the population in 2009, since it will have passed 6 years after 2003:
2009 - 2003 = 6
Therefore, we have:
P(2009) = 54000 + 2200 * 6
= 54000 + 13200
= 67200
So, the population in 2009 is 67200.
Andrew needs to replace a bolt that requires him to drill a hole 0.86 + - 0.001 inch.a.What are the upper and lower limits?b. Is he able to use a 55/64 drill bit?
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Hence he is able to use a 55/64 drill bit
Kuvo Graphing Linear Inequalities Sketch the wap each near in 12x-2
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Samantha, this is the solution to the exercise:
We need to plot the inequality:
x + 3y ≥ 3
x ≥ -3y + 3
y ≥ -1/3x + 1
Part 1: Predict the change from the parent function f(x)=x2 for each function. (Explain how it willlook different or transformed)1) g(x) = 4x22) h(x) = 5x23) j(x) = -3x2
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See explanation below
Explanation:Parent function = x²
For a parabola: y = ax²
If a approached a number greater than zero, the parabola becomes narrow. As a approaches, 0, the parabola widens.
1) g(x) = 4x²
Initial value of a in the parent function = 1
The value of a in this function = 4
4 > 1
This means the parabola will become narrow. And the parent function will be wider
2) h(x) = 5x²
a = 5
Initial value of a in the parent function = 1
5 > 1
The parabola will be more narrow than that of 4x².
3) j(x) = -3x²
a = -3
In this case, a is less than zero.
When a is greater than zero, the parabola opens upward. When it is less than zero, it opens downward.
From the graph, we will see the parabola opens downward.
The vertex in each of the graph remains the same.
Find and graph a 2nd degree polynomial function whose graph contains the point (0,-9)
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Okay, here we have this:
Considering the provided information, we are going to find the requested polynomial function, so we obtain the following:
Then let us remember that a second degree polynomial has the following form:
f(x)=ax^2+bx+c, where a, b and c are integers.
So replacing with the given point we get:
-9=a(0)^2+b(0)+c
-9=0+0+c
-9=c
Therefore any polynomial of the form ax^2+bx+c with c=-9 will pass through the given point, so if we take for example a=1, b=1 and c=-9, we obtain the following graph:
f(x)=1x^2+1x-9:
how do i find thr slope of the line for y=4/5x-3
Answers
Answer
Slope = m = (4/5)
Explanation
The slope and y-intercept form of the equation of a straight line is given as
y = mx + c
where
y = y-coordinate of a point on the line.
m = slope of the line.
x = x-coordinate of the point on the line whose y-coordinate is y.
c = y-intercept of the line.
So, comparing
y = (4/5)x - 3
with
y = mx + c
We can see that
Slope = m = (4/5)
y-intercept = c = -3
Hope this Helps!!!
Shapes ABCDE and JKLMN are similar, but not necessarily drawn to scale.What is the value of q and r
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As the figure ar similar, we have that
[tex]\frac{20}{8}=\frac{q}{7}=\frac{r}{5}[/tex]Therefore
[tex]q=7\cdot\frac{20}{8}=\frac{35}{2}=17.5[/tex]and
[tex]r=5\cdot\frac{20}{8}=\frac{25}{2}=12.5[/tex]so the answer is the second option
Height = -16x^2 +71x+29 Where X is the number of seconds after the baseball is thrown from the window
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The height function with respect to time has been given to be:
[tex]\begin{gathered} H(x)=-16x^2+71x+29 \\ \text{where x represents time(s}econds) \end{gathered}[/tex]a)
At x = 2secs, the height of the ball is:
[tex]\begin{gathered} H(2)=-16(2)^2+71(2)+29 \\ H(2)=-16(4)+142+29 \\ H(2)=-64+142+29 \\ H(2)=107feet \end{gathered}[/tex]Hence, at 2 seconds, the height of the ball is 107 feet
b)
At x = 4secs, the height of the ball is:
[tex]\begin{gathered} H(4)=-16(4)^2+71(4)+29 \\ H(4)=-16(16)+284+29 \\ H(4)=-256+284+29 \\ H(4)=57\text{ f}eet \end{gathered}[/tex]Hence, at 4 seconds, the height of the ball is 57 feet
what is 1+1???? I can't figure it out. someone said it is 12 but i think it is 11
Answers
Answer:
bro its 10
Step-by-step explanation:
1 + 1 = 10 (I using another number system, guess what)
Answer: 2
Step-by-step explanation: jfb23-_+_)_)_)O+_)(*&^%^&*()(*&^T
?"+*&*()_{"
}+_)(&^&)_+} This is how I got it
Hello! I need help solving this math practice problem. Thank you so much. I will include a picture of the question ! Thank you!
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Recall that the inequality:
[tex]x\ge2[/tex]as a statement is:
"All numbers greater than or equal to 2".
Therefore, the given set in the number line is:
The above set in interval notation is:
[tex]\lbrack2,\infty).[/tex]Answer:
[tex]\mleft\lbrace x|x\ge2\mright\rbrace=\lbrack2,\infty).[/tex]