Finding the mode and range of a data set Each day, Kaitlin records the number of news articles she reads. Here are her results for the last eight days. 7, 3, 8, 5, 7,7,7,8 Find the mode and the range for the data. Mode: Range: X 5 ?
Answers
Explanation:
The set of values are given below as
[tex]7,3,8,5,7,7,7,8[/tex]Mode:
This the data that occurs highest or the dat that has the highest frequency
Range:
The is the difference between the lowest val and the highest value
[tex]Range=highest-lowest[/tex]Hence,
The final answers are
[tex]\begin{gathered} mode=7(it\text{ occurs 4 times\rparen} \\ range=8-3=5 \end{gathered}[/tex]Hence,
The final answer is
[tex]\begin{gathered} mode=7 \\ Range=5 \end{gathered}[/tex]Related Questions
The first year shown the number of students per teacher fell below 16 was
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Using the y axis, we want to find when it goes below 16
The x value when y is less than 16 for the first time is 2002
The cost of a pair of skis to a store owner was $700, and she sold the pair of skis for $1020.Step 3 of 3: What was her percent of profit based on selling price? Follow the problem-solving process and round your answer to thenearest hundredth if necessary.
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Answer:
Explanation:
• The ,cost price ,of the pair of skis = $700
,• The ,selling price ,of the pair of skis = $1020
To calculate the percentage of profit, use the formula below:
[tex]\text{Percent of Profit=}\frac{Selling\text{ Price-Cost Price}}{\text{Selling Price}}\times\frac{100}{1}[/tex]Substitute the given values:
[tex]\text{Percent of Profit=}\frac{1020\text{-7}00}{\text{7}00}\times\frac{100}{1}\text{=}\frac{320}{\text{7}00}\times\frac{100}{1}=45.71\%[/tex]The percentage profit is % (correct to the nearest hundredth).
A glacier in Republica was observed to advance 22inches in a 15 minute period. At that rate, how many feet will the glacier advance in one year?
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To fins the rate in feet/year we must change first the measurements to the units required
inches to feat
minutes to years
[tex]22in\cdot\frac{1ft}{12in}=\frac{11}{6}ft[/tex][tex]15\min \cdot\frac{1h}{60\min}\cdot\frac{1day}{24h}\cdot\frac{1year}{365\text{days}}=\frac{1}{35040}\text{years}[/tex]to find the rate divide the distance over the time
[tex]\frac{\frac{11}{6}ft}{\frac{1}{35040}\text{year}}=\frac{11\cdot35040ft}{6\text{year}}=\frac{385440}{6}=\frac{64240ft}{\text{year}}[/tex]Х о 12 3 4 у -6 1 8 15 22what is the slope intercept form
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20 ping pong balls are numbered 1-20, with no repitition of any numbers. What is the probability of selecting one ball that is either odd or less than 5?
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given 20 ping pong balls
numbered 1-20
odd numbers = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19
total odd numbers = 10
numbers less than 5 = 1, 2, 3, 4
total numbers less than 5 = 4
since 1 and 3 are in both sides,
total number of porbabilities
= 10 + 4 - 2
= 12
the probability of selecting one ball
= 12/20
= 3/5
= 0.6
therefore the probabilty of selecting one ball that is either odd or less than 5 = 0.6
help meeeee pleaseeeee!!!
thank you
Answers
Step-by-step explanation:
what is the problem ?
first you need to put "1" in place of the x and calculate, and then you need to put "2" in place of the x and calculate.
just simple calculation !
(a)
R(1) = 1000×1² / (1² + 4) = 1000 / 5 = 200
$200 million
(b)
R(2) = 1000×2² / (2² + 4) = 4000 / 8 = 500
$500 million
there ! that's all that was needed.
Write a numerical expression for the word expression.The product of 2 groups of 7
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The given statement is the product of 2 groups of 7.
This statement can be expressed as
[tex]7\cdot7[/tex]Where each seven represent one group.
A spinner is divided into 5 equally sized segments colored blue, green, black, red, and yellow. Suppose you spin the wheel once and then spin it again. What is the probability of landing on the color red both times? Give your answer as an exact fraction and reduce the fraction as much as possible.
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The probability of landing in any colour is:
[tex]\frac{1}{5}[/tex]so if we want to get a red, both times, we to multiply 1/5 twice
[tex]\frac{1}{5}\cdot\frac{1}{5}=\frac{1}{25}[/tex]so, the probability of getting red twice is 1/25
A business could not collect $5,000 that it was owed. The total owed to the business was $100,000. What fraction of the total was not collected? (Express As Fraction)
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Total owed to the business = $100,000
amount that could not be collected = $5000
Fraction of total not collected
[tex]\text{fraction not collected=}\frac{5000}{100000}=\frac{5}{100}=\frac{1}{20}[/tex]The ratio of a quarterback's completed passes to attempted passes is 5 to 8. If he attempted 16 passes, find how many passes he completed. Round to the nearest whole number if necessary.
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The ratio of a quarterback's completed passes to attempted passes is 5 to 8. If he attempted 16 passes, find how many passes he completed. Round to the nearest whole number if necessary.
Let
x -----> number of quarterback's completed passes
y -----> number of quarterback's attempted passes
so
x/y=5/8 -----> x=(5/8)*y -----> equation A
y=16 -----> equation B
substitute equation B in equation A
x=(5/8)*16
x=10
therefore
the answer is 10 completed passesHaley spent 1/2 oven hour playing on her phone that used up 1/9 of her battery how long would she have to play on her phone to use the entire battery
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1/2 hour playing -- 1/9 battery
1 hour playing -- 2/9 battery
1 1/2 hours playing --- 3/9 battery
2 hours playing ------ 4/9 battery
2 1/2 hours playing ---- 5/9 battery
3 hours playing ----- 6/9 battery
3 1/2 hours playing --- 7/9 battery
4 hours playing -----8/9 battery
4 1/2 hours playing ---- 9/9 battery
9/9 represent the entire battery so che can play 4.5 hours on her phone
it can be represented into a fraction as
[tex]4.5=4\frac{1}{2}=\frac{9}{2}[/tex]* #3: Write the mixed number shown below as a decimal. 6 3/4
Answers
Answer:
6.75
Hope it helps!
Let me know if its wrong
There is a raffle with 250 tickets. One ticket will win a $320 prize, one ticket will win a $240 prize, one ticket will win a $180 prize, one ticket will win a $100 prize, and the remaining tickets will win nothing. If you have a ticket, what is the expected payoff
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Given that: There is a raffle with 250 tickets. One ticket will win a $320 prize, one ticket will win a $240 prize, one ticket will win a $180 prize, one ticket will win a $100 prize, and the remaining tickets will win nothing.
The expected payoff will be:
[tex]\begin{gathered} EV=\frac{1}{250}(320)+\frac{1}{250}(240)+\frac{1}{250}(180)+\frac{1}{250}(100)+\frac{246}{250}(0) \\ EV=\frac{320+240+180+100}{250} \\ EV=\frac{840}{250} \\ EV=3.36 \end{gathered}[/tex]So the expected payoff will be $3.36.
select the ordered pair that represent solutions of the system of inequalities A) (-1,3)B) (0,0)C) (-4,1)D) (0,2)E) (-5,0)F) ( -4,4)G) (-5,3)H) (-6,5)please answer fast
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From the graph, the solution of the graph is given by the area of intersection of the two functions. The are under the area of the two functions are (-4, 1), (-5, 0), (-4, 4)
Solve the system of equation by the elimination method {1/3x+1/2y=1/2{1/6x-1/3y=5/6(x,y)=(_, _)
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Solution
- The solution steps to solve the system of equations by elimination is given below:
[tex]\begin{gathered} \frac{x}{3}+\frac{y}{2}=\frac{1}{2}\text{ \lparen Equation 1\rparen} \\ \\ \frac{x}{6}-\frac{y}{3}=\frac{5}{6}\text{ \lparen Equation 2\rparen} \\ \\ \text{ Multiply Equation 2 by 2} \\ 2\times(\frac{x}{6}-\frac{y}{3})=\frac{5}{6}\times2 \\ \\ \frac{x}{3}-\frac{2y}{3}=\frac{5}{3}\text{ \lparen Equation 3\rparen} \\ \\ \\ \text{ Now, }\frac{x}{3}\text{ is common to both Equations 1 and 3.} \\ \\ \text{ We can therefore subtract both equations to eliminate }x. \\ \text{ We have:} \\ \text{ Equation 1 }-\text{ Equation 3} \\ \\ \frac{x}{3}+\frac{y}{2}-(\frac{x}{3}-\frac{2y}{3})=\frac{1}{2}-\frac{5}{3} \\ \\ \frac{x}{3}-\frac{x}{3}+\frac{y}{2}+\frac{2y}{3}=\frac{1}{2}-\frac{5}{3}=\frac{3}{6}-\frac{10}{6} \\ \\ \frac{y}{2}+\frac{2y}{3}=-\frac{7}{6} \\ \\ \frac{3y}{6}+\frac{4y}{6}=-\frac{7}{6} \\ \\ \frac{7y}{6}=-\frac{7}{6} \\ \\ \therefore y=-1 \\ \\ \text{ Substitute the value of }y\text{ into any of the equations, we have:} \\ \frac{1}{3}x+\frac{1}{2}y=\frac{1}{2} \\ \frac{1}{3}x+\frac{1}{2}(-1)=\frac{1}{2} \\ \\ \frac{1}{3}x=\frac{1}{2}+\frac{1}{2} \\ \\ \frac{1}{3}x=1 \\ \\ \therefore x=3 \end{gathered}[/tex]Final Answer
The answer is:
[tex]\begin{gathered} x=3,y=-1 \\ \\ \therefore(x,y)=(3,-1) \end{gathered}[/tex]Find the measures of the sine and cosine of the following triangles
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Let x be the side opposite to angle 62 degrees
Let y be the adjacent angle.
The sine of the angle is given as follows:
[tex]\begin{gathered} \sin62=\frac{Opposite}{Hypotenuse}=\frac{x}{10} \\ \end{gathered}[/tex]The cosine is given as:
[tex]\cos62=\frac{Adjacent}{Hypotenuse}=\frac{y}{10}[/tex]e) A client takes 1 1/2 tablets of medication three times per day for 4 days. How many tablets will the clienthave taken at the end of four days? Explain how your arrived at your answer.
Answers
Given
Client takes 1 1/2 of medication one time
Find
how many tablets will client have been taken at the end of 4 days.
Explanation
Client takes medication one time = 1 1/2
Client takes medication three times =
[tex]\begin{gathered} 1\frac{1}{2}\times3 \\ \frac{3}{2}\times3 \\ \frac{9}{2} \end{gathered}[/tex]medication for 4 days =
[tex]\begin{gathered} \frac{9}{2}\times4 \\ 18 \end{gathered}[/tex]Final Answer
The client takes 18 tablets at the end of four days.
In the accompanying regular pentagonal prism, suppose that each base edge measures 7 in. and that the apothem of the base measures 4.8 in. The altitude of the prism measures 10 in.A regular pentagonal prism and a pentagon are shown side by side. The pentagon contains a labeled segment and angle.The prism contains a horizontal top and bottom and vertical sides. The front left face and front right face meet the bottom base at right angles.The pentagon is labeled "Base".A line segment starts in the center of the pentagon, travels down vertically, and ends at the edge. The segment is labeled a.The vertical segment forms a right angle with the edge.(a)Find the lateral area (in square inches) of the prism.in2(b)Find the total area (in square inches) of the prism.in2(c)Find the volume (in cubic inches) of the prism.in3
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To determine the lateral area of the prism;
[tex]Lateral\text{ area=perimeter of the base}\times height[/tex][tex]Lateral\text{ area=5\lparen7\rparen }\times10=350in^2[/tex]To determine total area of the prism;
[tex]Total\text{ area=2\lparen area of base\rparen+Lateral area}[/tex][tex]\begin{gathered} Total\text{ area of the prism=2\lparen}\frac{1}{2}\times perimeter\text{ of the base}\times apotherm\text{\rparen+350} \\ \end{gathered}[/tex][tex]\begin{gathered} Total\text{ area of the prism=2\lparen}\frac{1}{2}\times5(7)\times4.8\text{\rparen+380=168+350=518in}^2 \\ \end{gathered}[/tex]To determine the volume of the prism;
[tex]Volume\text{ = base area }\times height[/tex][tex]Volume=\frac{1}{2}\times5(7)\times4.8\times10=840in^3[/tex]Hence
Given that angle A lies in Quadrant III and sin(A)= −17/19, evaluate cos(A).
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As we know;
[tex]sin^2(x)+cos^2(x)=1[/tex]We will use this equality. We take the square of the sine of the given angle and subtract it from [tex]1[/tex].
[tex]sin^2(A)=(-\frac{17}{19} )^2=\frac{289}{361}[/tex][tex]sin^2(A)+cos^2(A)=1[/tex][tex]sin^2(A)=1-cos^2(A)[/tex][tex]\frac{289}{361}=1-cos^2(A)[/tex][tex]cos^2(A)=1-\frac{289}{361} =\frac{72}{361}[/tex][tex]\sqrt{cos^2(A)} =cos(A)[/tex][tex]\sqrt{\frac{72}{361} }=\frac{6\sqrt{2} }{19}[/tex]In the third region the sign of cosines is negative. Therefore, our correct answer should be as follows;
[tex]cos(A)=-\frac{6\sqrt{2} }{19}[/tex]Tim and Kevin each sold candies and peanuts for a school fund-raiser. Tim sold 16 boxes of candies and 4 boxes of peanuts and earned $132. Kevin sold 6 boxes of peanuts and 20 boxes of candies and earned $190. Find the cost of each. Cost of a box of candy. Cost of a box of peanuts.
Answers
We have the following:
let x cost of a box of candy
let y cost of a box of peanuts
[tex]\begin{gathered} \text{ Tim} \\ 16x+4y=132 \\ \text{ Kevin} \\ 20x+6y=190 \end{gathered}[/tex]resolving the system of equations:
[tex]\begin{gathered} 20x+6y=190 \\ 16x+4y=132\Rightarrow4y=132-16x\Rightarrow y=\frac{132-16x}{4} \\ \text{replacing:} \\ 20x+6\cdot(\frac{132-16x}{4})=190 \\ 20x+198-24x=190 \\ -4x=190-198 \\ x=\frac{-8}{-4} \\ x=2 \end{gathered}[/tex]now, for y
[tex]\begin{gathered} y=\frac{132\cdot16\cdot2}{4} \\ y=25 \end{gathered}[/tex]Therefore the cost of the box of candy is $ 2 and the cost of the box of peanuts is $ 25
put the numbers in order from least to greatest2.3,12/5,5/2,2.2,21/10
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Express the fraction in terms of decimal.
[tex]\frac{12}{5}=2.4[/tex][tex]\frac{5}{2}=2.5[/tex][tex]\frac{21}{10}=2.1[/tex]The numbers are,
2.3, 2.4, 2.5, 2.2, 2.1.
Now we arrange the number from least to greatest.
[tex]2.1,2.2,2.3,2.4,2.5[/tex]So answer is,
[tex]\frac{21}{10},2.2,2.3,\frac{12}{5},\frac{5}{2}[/tex]12. Jimmy is paid $14.50 per hour for a regular forty-hour work week and 5 point time and a half for any hour worked over 40. This pas week, Jimmy earned $754.00 in total pay. How many hours of overtime did Jimmy work?
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Jimmy is paid $14.50 per hour for a regular forty-hour work week and 5 point time and a half for any hour worked over 40. This pas week, Jimmy earned $754.00 in total pay. How many hours of overtime did Jimmy work?
Let
x -----> the total hours worked
we have that
$14.50 --------> 40 hours
5.5($14.50) -------> > 40 hours
so
754=14.50*40+5.5(14.50)x
solve for x
754=580+79.75x
79.75x=754-580
79.75x=174
x=2.2 hours
What is the solution of the system of equations? Explain.18x+15-y=05y=90x+12
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The given system is
[tex]\begin{cases}18x+15-y=0 \\ 5y=90x+12\end{cases}[/tex]First, we multiply the first equation by 5.
[tex]\begin{cases}90x+75-5y=0 \\ 5y=90x+12\end{cases}[/tex]Then, we combine the equations
[tex]\begin{gathered} 90x+75-5y+5y=90x+12 \\ 90x+75=90x+12 \\ 75=12 \end{gathered}[/tex]Given that the result is not true (75 is not equal to 12), we can deduct that the system has no solutions.
help meeeee pleaseeeee!!!
thank you
Answers
Answer:
(f o g) = 464
Step-by-step explanation:
f(x) = x² - 3x + 4; g(x) = -5x
(f o g)(4) = f(g(4))
f(g(4)) = -5(4) = -20
f(g(4)) = (-20)² - 3(-20) + 4
f(g(4)) = 464
I hope this helps!
How do I solve:7/8+y= -1/8
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We are given the following equation
[tex]\frac{7}{8}+y=-\frac{1}{8}[/tex]Let us solve the equation for variable y
Our goal is to separate out the variable y
Subtract 7/8 from both sides of the equation.
[tex]\begin{gathered} \frac{7}{8}-\frac{7}{8}+y=-\frac{1}{8}-\frac{7}{8} \\ y=-\frac{1}{8}-\frac{7}{8} \end{gathered}[/tex]Since the denominators of the two fractions are the same, simply add the numerators.
[tex]\begin{gathered} y=-\frac{1}{8}-\frac{7}{8} \\ y=\frac{-1-7}{8} \\ y=\frac{-8}{8} \\ y=-1 \end{gathered}[/tex]Therefore, the value of y is -1
giving the figure below, what is the measure of angle JKL
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The measure of < JKL = 25+25 = 50 degrees.
Angle JOK = 360 -230 = 130 degrees , (where O is the center of the circle)
< OLK = < OJK = 90 degrees ( tangent to a circle)
< LOK = < JOK = 180 - (90+65) = 180 - 155 = 25 degrees
The solution is: < JKL = 25 +25 = 50 degrees
In the figure, ∆ABD ≅ ∆CBD by Angle-Side-Angle (ASA). Which segments are congruent by CPCTC? BC ≅ AD CB ≅ AB AB ≅ CD DB ≅ DC
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By CPCTC this is the only valid answer:
CB ≅ AB
Another statement should be AD≅ CD
Copper has a density of 4.44 g/cm3. What is the volume of 2.78 g of copper?
60 points please help
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The volume of 2.78 g of copper is 0.626 [tex]cm^{3}[/tex].
According to the question,
We have the following information:
Density of cooper = 4.44 [tex]g/cm^{3}[/tex]
Mass of copper = 2.78 g
We know that the following formula is used to find the density of any material:
Density = Mass/volume
Let's denote the volume of copper be V.
Now, putting the values of mass and density here:
4.44 = 2.78/V
V = 2.78/4.44
V = 0.626 [tex]cm^{3}[/tex]
(Note that the units if mass, volume and density are written with the numbers. For example, in this case, the unit of mass is grams, the unit of volume is [tex]cm^{3}[/tex].)
Hence, the volume of the copper is 0.626 [tex]cm^{3}[/tex].
To know more about volume here
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Here's a graph of a linear function. Write theequation that describes that function.Express it in slope-intercept form.Enter the correct answer.000DONEClear allĐOO
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Find the solution 5(x-9)+3=5x-42A) x=9B) x=-9C) Infinite SolutionsD) No Solutions
Answers
Answer:
C. Infinite Solutions
Explanation:
Given the equation
[tex]5\mleft(x-9\mright)+3=5x-42[/tex]First, open the bracket
[tex]\begin{gathered} 5x-45+3=5x-42 \\ 5x-42=5x-42 \end{gathered}[/tex]Since the left-hand side equals the right-hand side, the system has Infinite Solutions.