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elena-s [515]
4 years ago
11

I am a graph that records each piece of data above a number line

Mathematics
2 answers:
dimaraw [331]4 years ago
7 0
Ufudicucuvufigiviififig
kozerog [31]4 years ago
6 0

Answer:

Line plot

Step-by-step explanation:

Consider the provided information.

I am a graph that records each piece of data above a number line

Line plot is a graph that uses marks to record each piece of  given data above the number line.

It shows the frequency of data along a number line. Remember it is different from line graph.

The example of line plot is shown in figure below.

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If jake has 15 cookies and gives his 5 friends an even amount,how many will his friends receive?
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Answer:3

Step-by-step explanation:if you divide 3 and 15 is 5 so the answer is 3! Please give a lot of points!!!

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Help with algebra 2 I can’t figure out how to solve the problem .
defon

Answer: D

Step-by-step explanation:

Using the equation plot in the coordinates of the points to see if they fit the equation.

5x-3y < 30  

A)  5(-3) -3(5) < 30

   -15 -15 < 30

    -30 < 30   This is true

B)  5(0) -3(0) < 30

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C)  5(-5) - 3(3) < 30

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3 years ago
Louise sold 873 copies of a book she wrote for $10.99 each. If she earns 52% profit from each book, about how much profit will s
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CALCULUS EXPERT WANTED
Vlad [161]

a. Use the mean value theorem. 16 falls between 12 and 20, so

v'(16)\approx\dfrac{v(20)-v(12)}{20-12}=\dfrac{240-200}8=5

(Don't forget your units - 5 m/min^2)

b. v(t) gives the Johanna's velocity at time t. The magnitude of her velocity, or speed, is |v(t)|. Integrating this would tell us the total distance she has traveled whilst jogging.

The Riemann sum approximates the integral as

\displaystyle\int_0^{40}|v(t)|\,\mathrm dt=12\cdot200+8\cdot240+4\cdot220+16\cdot150=7600

If you're not sure how this is derived: we're given 5 sample points, so we can cut the interval [0, 40] into 4 subintervals. The lengths of each subinterval are 12, 8, 4, and 16 (the distances between each sample point), and the height of the rectangle approximating the area under the plot of |v(t)| is determined by the value of v(t) at each sample point, 200, 240, |-220| = 220, and 150.

c. Bob's velocity is given by B(t), so his acceleration is given by B'(t). We have

B'(t)=3t^2-12t

and at t=5 his acceleration is B'(5)=15 m/min^2.

d. Bob's average velocity over [0, 10] is given by the difference quotient,

\dfrac{B(10)-B(0)}{10-0}=\dfrac{700-300}{10}=40 m/min

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9/72


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