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DanielleElmas [232]
3 years ago
7

19, 25, 31, 19, 34, 22, 31, 34Mean:Median:Mode:Range:​

Mathematics
2 answers:
Anon25 [30]3 years ago
8 0

Answer:

mean: 27

median: 27

mode: 19, 31, 34

range: im very bad at finding this srry

Step-by-step explanation:

pls tell me if im wrong

Ksju [112]3 years ago
3 0
Mean- 26.875
Median- 28
Mode- 34,19,31
Range- 15
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15 is 30% of what number?
dangina [55]
#1---> let the number you are trying to find be x for the following questions

30%*x=15
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#2
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#3
(4/32)*100
 is the same as (1/8)*100= 12.5%

#4
1%* x=11
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If you take a number, multiply it by 3, then add 2, you could represent that phrase with the function:
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Step-by-step explanation:

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2 years ago
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Write the given second order equation as its equivalent system of first order equations. u′′−5u′−4u=1.5sin(3t),u(1)=1,u′(1)=2.5
lakkis [162]

Answer:

hi your question options is not available but attached to the answer is a complete question with the question options that you seek answer to

Answer:  v = 5v + 4u + 1.5sin(3t),

  • 0
  • 1
  •  4
  • 5
  • 0
  • 1.5sin(3t)
  • 1
  •  2.5

Step-by-step explanation:

u" - 5u' - 4u = 1.5sin(3t)        where u'(1) = 2.5   u(1) = 1

v represents the "velocity function"   i.e   v = u'(t)

As v = u'(t)

<em>u' = v</em>

since <em>u' = v </em>

v' = u"

v'  = 5u' + 4u + 1.5sin(3t)   ( given that u" - 5u' - 4u = 1.5sin(3t) )

    = 5v + 4u + 1.5sin(3t)  ( noting that v = u' )

so v' = 5v + 4u + 1.5sin(3t)

d/dt \left[\begin{array}{ccc}u&\\v&\\\end{array}\right]= \left[\begin{array}{ccc}0&1&\\4&5&\\\end{array}\right]  \left[\begin{array}{ccc}u&\\v&\\\end{array}\right] + \left[\begin{array}{ccc}0&\\1.5sin(3t)&\\\end{array}\right]

Given that u(1) = 1 and u'(1) = 2.5

since v = u'

v(1) = 2.5

note: the initial value for the vector valued function is given as

\left[\begin{array}{ccc}u(1)&\\v(1)\\\end{array}\right]  = \left[\begin{array}{ccc}1\\2.5\\\end{array}\right]

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On a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases in quadrant 2. It crosses the y-axis at (0, 0.25) and goes through (negative 1, 2).

On a coordinate plane, an exponential function approaches y = 0 in quadrant 3 and decreases into quadrant 4. It crosses the y-axis at (0, negative 0.25) and goes through (1, negative 2).

On a coordinate plane, an exponential funtion increases in quadrant 3 into quadrant 4 and approaches y = 0. It goes through (negative 1, negative 2) and crosses the y-axis at (0, negative 0.25).Step-by-step explanation:

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