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kifflom [539]
2 years ago
9

Suppose a normal distribution has a mean of 62 and a standard deviation of

Mathematics
1 answer:
tankabanditka [31]2 years ago
6 0

Answer:

66.8% or 0.66772

Step-by-step explanation:

0.66772

Given that :

Mean (m) = 62

Standard deviation (s) = 4

probability that a data value is between 57 and 65

P(57 < x < 65)

Using the relation :

(x - m) / s

P(x < 57) :

(57 - 62) / 4 = - 1.25

P(Z < - 1.25) = 0.10565 ( Z probability calculator)

P(x < 65) :

(65 - 62) / 4 = 0.75

P(Z < 0.75) = 0.77337 (Z probability calculator)

P(Z < 0.75) - P(Z < - 1.25)

0.77337 - 0.10565 = 0.66772

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Answer:

i think its c im not sure so it might be wrong im so sorry if its wrong

Step-by-step explanation:

6 0
3 years ago
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
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Judy is playing a card game. She needs to select two cards below that have a sum greater
Zielflug [23.3K]

Answer:

2\frac{2}{5}  + 2\frac{5}{8} \simeq 5

2\frac{2}{7}  + 2\frac{5}{8}  \simeq 5

Step-by-step explanation:

Mixed fraction a\frac{b}{c} can be converted into like or unlike fractions.

Here, a\frac{b}{c} =  \frac{ac +b}{b}

So, convert all the given fractions as follows:

2\frac{2}{5} =  \frac{2(5) +2}{5} = \frac{12}{5}  =2.4

2\frac{2}{7} =  \frac{2(7) +2}{7} = \frac{16}{7}  =2.4

1\frac{7}{8} =  \frac{1(8) +7}{8} = \frac{15}{8}  =1.8

2\frac{1}{10} =  \frac{2(10) + 1}{10} = \frac{21}{10}  =2.1

2\frac{5}{8} =  \frac{2(8) +5}{8} = \frac{21}{8}  =2.6

Here, No Two fractions on addition gives a sum greater than 5.

But an approximate sum of 5 is given by adding 2.4 + 2.6 = 5

⇒2\frac{2}{5}  + 2\frac{5}{8} \simeq 5

or  ⇒2\frac{2}{7}  + 2\frac{5}{8}  \simeq 5

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Answer:

I would help but the thing is there is no options so sorry man

Step-by-step explanation:

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The final number of outcomes would end up being 10 possible outcomes because you would do five and multiply it by two. And there would be a possibility of landing on any of ten outcomes. Hope I helped reassure you of your answer :)
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