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lina2011 [118]
3 years ago
15

Which of the following inequalities is correct?

Mathematics
2 answers:
Elza [17]3 years ago
7 0

Answer:6 should be right

Step-by-step explanation: since the absolute value of -6 is 6, it satisfies the equation

|-6|=6

6 is less than or EQUAL to 6

Elina [12.6K]3 years ago
3 0

Answer:

The last one

Step-by-step explanation:

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A random sample of 50 bottles is selected from the production line of a large manufacturing company. The mean weight of the cont
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The population is all bottles in the production line and the sample is the 50 bottles that were selected.

Step-by-step explanation:

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4 0
3 years ago
Pleeeaaasseeee help (questions in photo) will mark brainliest
KIM [24]

Answer:

a. 0.8727rad

b. 5.585rad

that should be correct, if not, i apologize T_T

5 0
2 years ago
Which is the best estimate of x based on rounding the constants and coefficients in the equation to the nearest integer 6.15 (x
artcher [175]
Your answer would be x = 12.41 

6.15 (x - 8.86) = 21.83
      *6.15 * 6.15

6.15x - 54.49 = 21.83
         + 54.49  + 54.49

6.15x = 76.32
/6.15    /6.15

x = 12.41
7 0
3 years ago
The questions are in the picture
o-na [289]

Answer:

I can't see the picture

Step-by-step explanation:

I can't see it

4 0
3 years ago
Read 2 more answers
Find the radius of convergence, then determine the interval of convergence
galben [10]

The radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series. This can be obtained by using ratio test.  

<h3>Find the radius of convergence R and the interval of convergence:</h3>

Ratio test is the test that is used to find the convergence of the given power series.  

First aₙ is noted and then aₙ₊₁ is noted.

For  ∑ aₙ,  aₙ and aₙ₊₁ is noted.

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = β

  • If β < 1, then the series converges
  • If β > 1, then the series diverges
  • If β = 1, then the series inconclusive

Here a_{k} = \frac{(x+2)^{k}}{\sqrt{k} }  and  a_{k+1} = \frac{(x+2)^{k+1}}{\sqrt{k+1} }

   

Now limit is taken,

\lim_{n \to \infty} |\frac{a_{n+1}}{a_{n} }| = \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }/\frac{(x+2)^{k} }{\sqrt{k} }|

= \lim_{n \to \infty} |\frac{(x+2)^{k+1} }{\sqrt{k+1} }\frac{\sqrt{k} }{(x+2)^{k}}|

= \lim_{n \to \infty} |{(x+2) } }{\sqrt{\frac{k}{k+1} } }}|

= |{x+2 }|\lim_{n \to \infty}}{\sqrt{\frac{k}{k+1} } }}

= |{x+2 }| < 1

- 1 < {x+2 } < 1

- 1 - 2 < x < 1 - 2

- 3 < x < - 1

 

We get that,

interval of convergence = (-3, -1)

radius of convergence R = 1

Hence the radius of convergence R is 1 and the interval of convergence is (-3, -1) for the given power series.

Learn more about radius of convergence here:

brainly.com/question/14394994

#SPJ1

5 0
1 year ago
Read 2 more answers
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