The coordinates of V’ in the image of TUV under a dilation centered at the origin with scale 5 followed by a translation 3 units right and 2 units down can be found by following the transformation sequence.
What are the coordinates of V’ in the image of TUV under a dilation centered at the origin with scale 5 followed by a translation 3 units right and 2 units down, given T(-1,-1), U (-1,2), and V(2,1)?The first transformation to be applied is the dilation with a scale of 5. This transformation enlarges the coordinates of the original points by a factor of 5. The coordinates of T, U, and V become (5(-1,-1)), (5(-1,2)), and (5(2,1)) respectively. The coordinates of the dilated points are (T’(-5,-5)), (U’(-5,10)), and (V’(10,5)).The next transformation to be applied is the translation 3 units right and 2 units down.This transformation adds 3 to the x-coordinate and subtracts 2 from the y-coordinate of each point. After the translation, the coordinates of T’, U’, and V’ become (T’(-2,-7)), (U’(-2,8)), and (V’(13,3)) respectively.Therefore, the coordinates of V’ in the image of TUV under a dilation centered at the origin with scale 5 followed by a translation 3 units right and 2 units down are (13,3).The coordinates of V' can be found by first applying the dilation with scale 5 to the original points, followed by a translation 3 units right and 2 units down. Using the coordinates of T, U, and V as starting points, the coordinates of the points after the dilation with scale 5 can be found. T(-1,-1) becomes (-5,-5), U(-1,2) becomes (-5,10), and V(2,1) becomes (10,5). Applying the translation of 3 units right and 2 units down to the new points, the coordinates of V' can be found. T(-5,-5) becomes (-2,-7), U(-5,10) becomes (-2,8), and V(10,5) becomes (13,3). Therefore, the coordinates of V' are (13,3).To learn more about a dilation and a translation refer to:
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A student missed 45 problems on a mathematics test and received a grade of 39%. If all the problems were of equal value, how many problems were on the test
There were 74 problems in the test.
Rounding-off percentageLet the student got total number of problems in the test to be X.
Percentage of correct answer = 39 %
⇒ 61% of X were incorrect
⇒61/100 x X = 45
⇒61X = 45x100
⇒X = 4500/61
⇒X=73.77
After rounding off 73.77, we get 74.
Hence, we can say that there were total of 74 problems in the test.
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Part b
calculate the perimeter of each quadrilateral and the ratio of the perimeters. round your answers to the hundredths place.
The perimeter of the quadrilateral illustrated is 30cm.
How to calculate the perimeter?The information is incomplete. Therefore, an overview will be given. It should be noted ha perimeter simply means the addition of the sides given.
Let's assume that that we've a rectangle with length and width of 10 and 5cm.
The perimeter will be:
= 2(l + w)
= 2(10 + 5)
= 30cm
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During a snowstorm, Madeline tracked the amount of snow on the ground. When the storm began, there were 3 inches of snow on the ground. Snow fell at a constant rate of 1 inch per hour until another 2 inches had fallen. The storm then stopped for 3 hours and then started again at a constant rate of 2 inches every 3 hours for the next 9 hours. When the storm stopped again, the sun came out and melted the snow for the next 2 hours at a constant rate of 3 inches per hour. Make a graph showing the inches of snow on the ground over time using the data that Madeline collected.
The graph depicts that at the end of the snowstorm, there was 5 inches of snow left. See the attached graph.
How many hours did the snow fall?It is not be noted that the total amount of hours which the snow fell is:
2 + 9 = 11 hours.
Although the graph depicts a peak of 14 hours, recall that there were three hours when the snow didn't fall.
While the maximum depth of the snow was 11 inches.
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A regular pair of gloves are $35. If its sale price is $24.50, what is the percent of disount?
Answer:
30% discount
Step-by-step explanation:
(35 - 24.50) / 35 * 100% = 30%
A model pirate ship uses the scale
5 inch: 20 meters. If the model is
50 inches long, how long is the
pirate ship?
Answer:
200 meters
Step-by-step explanation:
If the ratio is 5 inches to 20 meters that means for every 5 inches in the model there will be 20 meters on the real ship.
If the model is 50 inches long that means that there are 10 5 inch segments, multiply this 10 by the 20 meters in the ratio and you will get 200 meters as the final length for the ship.
Please help! It is due right now!!!
The total bill for dinner was $33.03 (including tax and a tip). If they paid a 20.2% tip, what was their bill before adding the tip?
Using proportions, it is found that their bill before adding the tip was of $27.48.
What is a proportion?A proportion is a fraction of a total amount, and the measures are related using a rule of three.
They paid a 20.2% tip, hence 120.2% = 1.202 of the price x is equals to $33.03, hence the bill before the tip is found as follows:
1.202x = 33.03
x = 33.03/1.202
x = $27.48.
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pls help asap............
See below for the solution to each question
Is the graph a function?Yes, the graph is a function.
This is because all x values have different y values
The domainThis is the set of input values of the graph.
From the graph, we have
x = 0 to x = 17
Hence, the domain is [0, 17]
The rangeThis is the set of output values of the graph.
From the graph, we have
y = 0 to y = 10
Hence, the range is [0, 10]
The maximumThis is the maximum point on the graph.
From the graph, we have
Maximum = (12, 10)
The minimumThis is the minimum point on the graph.
From the graph, we have
Minimum = (0, 0)
The increasing intervalsThese are the intervals where the y values increase as x increase.
From the graph, we have
Increasing intervals = (0, 5) ∪ (10, 12) ∪ (14, 15)
The decreasing intervalsThese are the intervals where the y values decrease as x increase.
From the graph, we have
Decreasing intervals = (7, 10) ∪ (12, 14) ∪ (15, 17)
The constant intervalsThese are the intervals where the y values remain unchanged as x changes.
From the graph, we have
Constant intervals = (5, 7)
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Solving inequalities( need help checking my answer)
Step 1. You need x isolated.
[tex]x < 13+7[/tex]
When you move a number from the left to right (and its addition or subtraction, the sign stays (< ≤ > and ≥ is signs).
x<20 or in words, x is less than 20, and cannot be 20.
Answer:
x < 20
Any number less than 20 is a solution
Step-by-step explanation:
x-7 < 13
The first step to solve this inequality is to add 7 to each side
x-7+7 < 13+7
x < 20
Any number less than 20 is a solution
Vic is standing on the ground at a point directly south of the base of the CN Tower and can see the top when looking at an angle of elevation of 61°. Dan is standing on the ground at a point directly west of the base of the tower and must look up at an angle of elevation of 72° in order to see the top. If the CN Tower is 553.3 m tall,how far apart are Vic and Dan to the nearest meter? Include a well-labeled diagram as part of your solution.
Vic and Dan are 2, 897m apart.
How to determine the distance
It is important to note that the distance between Vic and Dan is the base of CN
Let's say the distance to Dan is x
The distance to Vic is y
Using cosine ratio, we have
cos α = opposite / adjacent
α = 72°
opposite = 553. 3cm
Adjacent = x
cos 72° = [tex]\frac{553. 3}{x}[/tex]
Cross multiply
[tex]cos 72[/tex] × [tex]x[/tex] = [tex]553. 3[/tex]
[tex]0. 3090x= 553. 3[/tex]
[tex]x = \frac{553. 3}{0. 3090}[/tex]
[tex]x = 1, 790. 61[/tex] m
The distance to Vic is y
Using the cosine ratio, we have
[tex]cos 60 = \frac{553. 3}{y}[/tex]
Cross multiply
[tex]0. 5y = 553. 3[/tex]
[tex]y = \frac{553. 3}{0. 5}[/tex]
[tex]y = 1,106. 6[/tex]m
To determine how far apart Vic and Dan, we use = x + y
= 1790. 61 + 1106. 6
= 2, 897. 21m
= 2, 897m
Thus, Vic and Dan are 2, 897m apart.
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For how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1 2 \dots n$
The value of positive integers in the set are 16.
According to the statement
we have given that the there is a set of numbers from 1 to n and we have to find that the how many integers in this set. and there is one condition that the numbers in the set are less than or equal to 24.
So, For this purpose,
Since [tex]$1 + 2 + \cdots + n = \frac{n(n+1)}{2}$[/tex]
the condition is equivalent to having an integer value for [tex]$\frac{n!} {\frac{n(n+1)}{2}}$.[/tex]
This reduces, when [tex]$n\ge 1$[/tex], to having an integer value for [tex]$\frac{2(n-1)!}{n+1}$[/tex]
This fraction is an integer unless n+1 is an odd prime. There are 8 odd primes less than or equal to 24,
so there are 24-8 = 16.
So, The value of positive integers in the set are 16.
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PLEASE HELPPPoirnginrigniniervifire
Answer:
555 m
Step-by-step explanation:
Because the model should have identical proportions to the real-life tower, the ratio between the height and shadow length of both diagrams should be the same.
Height (model) Height (real)
------------------------- = ------------------------
Shadow (model) Shadow (real)
Math:
0.9 meters x
-------------------- = ---------------------- <----- Ratio
0.2 meters 123.3 meters
4.5 = x / 123.3 meters <----- Divide 0.2 from 0.9
555 meters = x <----- Multiply both sides by 123.3
A professor gave her students six essay questions from which she will select three for a test. A student has time to study for only three of these questions. What is the probability that, of the questions studied.
The probability that of the questions studied, all three were the test questions selected by the professor is 0.05.
How to find the probability?We can find the probability by dividing the total number of favorable events by the total number of events.
The total number of events is given by [tex]6C_{3}[/tex].
The total number of favorable events is given by [tex]3C_{3}[/tex].
The probability is given by = Favorable events/Total number of events
= [tex]\frac{3C_{3}}{6C_{3}}[/tex]
= [tex]\frac{1}{\frac{(6)(5)(4)}{(1)(2)(3)} }[/tex]
= [tex]\frac{1}{(5)(4) }[/tex]
= 0.05
The probability that all three questions are selected by the professor for the test is found to be 0.05.
Therefore, we have found that the probability that of the questions studied, all three were the test questions selected by the professor is 0.05.
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Disclaimer: The question was incomplete, the complete question is attached below.
Question: A professor gave her students six essay questions from which she will select three for a test. A student has time to study for only three of these six questions. What is the probability that, of the questions studied, all three were the test questions selected by the professor?
Three more than twice a number is five less than the square of the number.What is the number?
The denominator of a fraction exceeds numerator by 3. If the numerator is doubled and the denominator is increased by 14, then fraction becomes 2/3rd of the original fraction.
Answer:
The original fraction is 4/7Step-by-step explanation:
Let the fraction be x/y.
According to question we have the following equations.
The denominator of a fraction exceeds numerator by 3:
y = x + 3If the numerator is doubled and the denominator is increased by 14, then fraction becomes 2/3rd of the original fraction:
2x/(y + 14) = (2/3)*(x/y)Change the fraction as below and solve for y:
2x /(y + 14) = 2x/(3y) Nominators are samey + 14 = 3y Compare denominators2y = 14y = 7Find the value of x using the first equation:
7 = x + 3x = 7 - 3x = 4The fraction is:
x/y = 4/7Answer:
Original fraction = ⁴/₇
Step-by-step explanation:
Numerator: top of the fraction
Denominator: bottom of a fraction
Let x be the original numerator.
If the denominator of a fraction exceeds the numerator by 3:
[tex]\implies \dfrac{x}{x+3}[/tex]
If the numerator is doubled and the denominator is increased by 14, then fraction becomes 2/3rd of the original fraction:
[tex]\implies \dfrac{2x}{x+3+14}=\dfrac{2}{3}\left(\dfrac{x}{x+3}\right)[/tex]
[tex]\implies \dfrac{2x}{x+17}=\dfrac{2x}{3(x+3)}[/tex]
[tex]\implies \dfrac{2x}{x+17}=\dfrac{2x}{3x+9}[/tex]
Cross multiply:
[tex]\implies 2x(3x+9)=2x(x+17)[/tex]
Divide both sides by 2x:
[tex]\implies 3x+9=x+17[/tex]
Subtract x from both sides:
[tex]\implies 2x+9=17[/tex]
Subtract 9 from both sides:
[tex]\implies 2x=8[/tex]
Divide both sides by 2:
[tex]\implies x=4[/tex]
Substitute the found value of x into the original fraction:
[tex]\implies \dfrac{4}{4+3}=\dfrac{4}{7}[/tex]
Therefore, the original fraction is ⁴/₇.
The area of the base of the cone is 8 pi mm^2. what is the volume of the cone in terms of pi?
The volume of cone whose area of base is 8π[tex]mm^{2}[/tex] is 32π/3 [tex]mm^{3}[/tex].
Given area of base of the cone 8π[tex]mm^{2}[/tex].
We are required to find the volume of the cone.
We know that base of a cone used to be in circle so the area of base of cone is equal to π[tex]r^{2}[/tex].
Area=8π (given)
π[tex]r^{2}[/tex]=8π
[tex]r^{2}[/tex]=8
r=[tex]\sqrt{8}[/tex]
r=2[tex]\sqrt{2}[/tex]
Volume of cone=1/3*π[tex]r^{2} h[/tex]
=1/3*π[tex](2\sqrt{2} )^{2}[/tex]*4
=1/3*8*4π
=32π/3
(We are not required to put value of π so our answer will be 32π/3.)
Hence the volume of cone whose area of base is 8π is 32π/3.
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Question is incomplete as it should also include height of cone be 4 mm.
Help me with this please
Answer:
Step-by-step explanation:
PLEASE HELP IM STUCK
Answer:
#3
Step-by-step explanation:
To create a number line to represent x<= 0, draw a solid circle at 0 (because 0 is included), then draw an arrow extending to the left (because it's less than or equal to).
Instructions: State what additional information is required in order to know that the triangles in the image below are congruent for the reason given.
Reason: SAS Postulate
When two or more given triangles are congruent, this implies that the measures of their corresponding angles and sides are equal. So that the additional information required in the question is FG ≅ IJ.
When two or more given triangles are congruent, this implies that the measures of their corresponding angles and sides are equal. Thus the congruent relations of the triangles can be with respect to their angles or/ and sides.
The two given triangles in the question would be congruent with respect to the Side-Angle-Side (SAS) postulate if the condition below is true:
i.e FG ≅ IJ
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A circular table has a radius of 5cm. decorative trim is placed along the outside edge. how long is the trim?single line text.
If the radius of the circular table is 5 cm then the length of the trim required is 31.4 cm.
Given that the radius of the circular table is 5 cm.
What is the length of trim needed to decorate along the outside trim?
Circumference is the length of arc of the circle.It is also known as the perimeter of the circle.
Circumference of the circle=2πr in which r is the radius of the circle.
It is given that the trim is placed and decorated along the outside edge, so the perimeter of the circle must equal to the length of trim needed.
Length of trim needed to decorate the circular table=2πr
=2*π*5
=10π
=10*3.14
=31.4 cm.
Hence the length of the trim needed to decorate along the table is 31.4 cm.
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Two numbers are independently selected from the set of positive integers less than or equal to 5. What is the probability that the sum of the two numbers is greater than their product
There is a 9/25 chance that two numbers will be added together such that their total exceeds their product.
How to find out the probabilityThe set of positive numbers are:
{1,2,3,4,5}
We have to independently select two numbers from these sets.
We are trying to determine the likelihood that the sum is larger than the product of the integers.
When one of the chosen numbers is 1, the total will always be greater than the product because:
1*1=1 & 1+1=2
1*2=2 & 1+2=3
1*3=3 & 1+3=4
1*4=4 & 1+4=5
1*5=5 & 1+5=6
and so on.
If we select 2 as both numbers then,
2*2=4 & 2+2=4
Here sum and the product are equal.
If not, the product will be bigger than the sum.
Now, the first stage is to evaluate the total number of two combinations that are feasible:
We have 5 alternatives for the first number and 5 alternatives for the second number.
The product of the number of possibilities in each scenario yields the total number of combinations:
C= 5*5=25
The following combinations have a sum that is greater than the product :
1 and 1
1 and 2
1 and 3
1 and 4
1 and 5
2 and 1
3 and 1
4 and 1
5 and 1
So, we get 9 combinations.
So, Probability,p=9/25
Therefore it is concluded that the probability is 9/25.
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Wey wey wonders how much water it would take to fill her cup. she drops her pencil in her cup and notices that it only fits diagonally. the pencil is 17 centimeters long and the cup is 15 centimeters tall. how much water can the cup hold? use paper to help you with your thinking.
Answer:
240π cm³ (exact)
754 cm³ (approximate)
Step-by-step explanation:
I assume the shape of the cup is a cylinder. The height of the cylinder is 15 cm.
Think of a vertical rectangle that contains the vertical axes of the cylinder.
Two sides of the rectangle are diameters of the bases of the cylinder. The other two side are heights of the cylinder. The pencil is 17 cm long and is a diagonal of this rectangle. We can find the diameter of the base of the cylinder using the Pythagorean theorem.
In a rectangle that is 15 cm long, the diagonal measures 17 cm.
a² + b² = c²
15² + b² = 17²
b = 8
The diameter of the cup is 8 cm.
radius = 8 cm/2 = 4 cm
V = πr²h
V = π(4 cm)²(15 cm) = 240π cm³ (exact)
V = 754 cm³ (approximate)
Provide reasons for the statements.
Given: ∠1 and ∠3 are vertical angles.
Prove: ∠1 ≅ ∠3
(See image for more details on the equation, please.)
The reasons for the statement about the angles include:
3. Definition of linear pair.
4. Definition of supplementary angles.
5. Substitution.
6. Subtraction property of equality.
7. Definition of congruence.
How to illustrate the information?The angles are supplementary because the angles 1 and 2, 2 and 3 form a linear pair and thus because of the definition of linear pair, they are supplementary angles.
The sum of angles of 1 and 2, 2 and 3 is equal to 180 because these pair of angles are supplementary angles. Supplementary angles are those angles whose sum is 180 degrees.
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What is the following sum?
3/125x10,13 +3/27x10,13
O
○ 8x³y² (¹³√xy)
15x³y³ (³√xy )
O 15x5
O
15x³y² (2√xy)
O
○ 8x³y³ (³√xy)
[tex]\sqrt[3]{125x^{10}y^{13} } +\sqrt[3]{27x^{10}y^{13} }[/tex]=[tex]5\sqrt[3]{x^{10} y^{13} } +3\sqrt[3]{x^{10}y^{13} } = 8\sqrt[3]{x^{10}y^{13} }[/tex]
How to solve an expression?[tex]\sqrt[3]{125x^{10}y^{13} } +\sqrt[3]{27x^{10}y^{13} }[/tex]
Therefore,
125 = 5³
27 = 3³
Hence,
[tex]5\sqrt[3]{x^{10} y^{13} } +3\sqrt[3]{x^{10}y^{13} }[/tex]
Hence,
Both cube root are the same now,
Therefore, we have to add them together.
[tex]5\sqrt[3]{x^{10} y^{13} } +3\sqrt[3]{x^{10}y^{13} } = 8\sqrt[3]{x^{10}y^{13} }[/tex]
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The radius of circle A is 4.4 units.Which element of circle A has a measure of 27.65 units
Answer:
The circumference
Step-by-step explanation:
[tex]circumference \: = 2\pi \times radius \\ c = 2\pi(4.4) \\ c = 8.8\pi = 27.65[/tex]
The average height of students at UH from an SRS of 14 students gave a standard deviation of 2.5 feet. Construct a 95% confidence interval for the standard deviation of the height of students at UH. Assume normality for the data.
The 95% confidence interval for the standard deviation of the height of students at UH is given by; CI = (1.81, 4.03)
How to find the confidence interval for standard deviation?
The formula for the confidence interval for the standard deviation is given by the formula;
CI = √[(n - 1)s²/(χ²ₙ ₋ ₁, α/2)], √[(n - 1)s²/(χ²ₙ ₋ ₁, (1 - α)/2)]
We are given;
Sample size; n = 14
D F = n - 1 = 14 - 1 = 13
Standard Deviation; s = 2.5
Confidence Level; CL = 95% = 0.95
Significance level; α = 1 - 0.95 = 0.05
Thus, using Chi-square distribution table online we have;
χ²₁₃, ₀.₀₂₅ = 24.736
(χ²₁₃, ₀.₉₇₅) = 5.01
Now, the 95% confidence interval for the standard deviation of the height of students at UH is given by :-
CI = √[(13 * 2.5²/(24.736)], √[(13 * 2.5²/(5.01)]
CI = (1.81, 4.03)
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Jaun rides the bus to school each day he always arrives at his bus stop on time but his bus is late 80% of the time
The correct probability that Juan's bus is going to be late every week next week is 20 percent.
How to solve for the probabilityWe have the total number in the stimulation on to be from 0 to 9
On the fact that it would be late, the number ranges from 2 to 9
Hence the fact that it would be late would be
2/10
= 0.2
0.2 is also the same as 20 percent.
Complete questionJuan rides the bus to school each day. He always arrives at his bus stop on time, but his bus is late 80% of the time. Juan runs a simulation to model this using a random number generator. He assigns these digits to the possible outcomes for each day of the week:
• Let 0 and 1 = bus is on time
• Let 2, 3, 4, 5, 6, 7, 8, and 9 = bus is late
The table shows the results of the simulation.
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Six is called a perfect number because its factors (not including 6) add up to itself, i.e. 1 + 2 + 3 = 6. find all the perfect numbers between 20 and 30
The calculator shows the result that Enrique got after evaluating the expression 56 + 7 × (34 – 17) – 16 . He checked his work by rounding all of the values to the nearest ten and comparing it to the calculator result. Which is true regarding his estimate and the accuracy of his calculator result?
Answer:
The estimated result is 680, which suggests the calculator is incorrect.
answer choices would have helped, hope this helps :)
A taxi company charges a $2.00 base fare plus an additional fare based on a per-mile rate and a per-minute rate. Ryan's first taxi ride was 3.0 miles, took 7 minutes, and cost $8.50. His second taxi ride was 7.0 miles, took 14 minutes, and cost $16.00. If his third taxi ride took 10 minutes and cost $13.50, approximately how many miles was the third taxi ride
A $2.00 base fare charge, and per-mile, and per-minute rates related as follows; 3•x + 7•y = 6.5 and 7•x + 14•y = 14, give the distance traveled in the third ride as 4.5 miles
How can the length of the third ride be calculated?The base fare = $2.00
Let x represent the per-mile rate, and let y represent the per-minute rate, we have;
The cost of Ryan's first taxi ride = $8.50
Distance traveled in the first ride = 3.0 miles
Time taken during the first ride = 7 minutes
Therefore;
2 + 3•x + 7•y = 8.5
Which gives;
3•x + 7•y = 8.5 - 2 = 6.5
3•x + 7•y = 6.5...(1)Distance traveled in the second ride = 7.0 miles
Duration of the second ride = 14 minutes
The second ride cost = $16.00
Therefore;
2 + 7•x + 14•y = 16
Which gives;
7•x + 14•y = 16 - 2 = 14
7•x + 14•y = 14...(2)Solving the above simultaneous equations by multiplying equation (1) by 2 then subtracting the result from equation (2) gives;
(7•x + 14•y) - 2 × (3•x + 7•y) = 14 - 2×6.5 = 1
x = 1
The per-mile rate, x = $13•x + 7•y = 6.5
7•y = 6.5 - 3•x
7•y = 6.5 - 3×1 = 3.5
y = 3.5/7 = 1/2 = 0.5
The per-minute rate, y = $0.5Duration of the third taxi ride = 10 minutes
Cost of the third ride = $13.50
Therefore;
2 + 1×a + 10×y = $13.50
Where a is the distance traveled during the third ride, we have;
2 + 1×a + 10×0.5 = $13.50
2 + a + 5 = 13.5
a = 13.5 - 2 - 7 = 4.5
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Suppose in a class of 60 students 5 have no siblings, 26 have one sibling, 14 have two siblings, and 15 have three siblings. Calculate the relative frequency of students who have three siblings. (please express as a percentage)
The relative frequency of students who have three siblings is 25% given that there are 60 students in a class in which 5 have no siblings, 26 have one sibling, 14 have two siblings and 15 have three siblings. This can be obtained by using the formula for relative frequency.
What is the relative frequency of students who have three siblings?Given that,
total number of students in the class = 60
number of students who have no sibling = 5
number of students who have 1 sibling = 26
number of students who have 2 sibling = 14
number of students who have 3 sibling = 15
Formula for relative frequency = f/n, where f is the number of times the data occurred, n is the total number of frequencies.
Therefore,
relative frequency of students who have three siblings = 15/60 = 0.25
In percentage ⇒ 0.25 × 100 = 25%
Hence the relative frequency of students who have three siblings is 25% given that there are 60 students in a class in which 5 have no siblings, 26 have one sibling, 14 have two siblings and 15 have three siblings.
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