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jekas [21]
3 years ago
15

The number of lightening strikes on a square kilometer of open ground in a year has mean 5 and standard deviation 2.1. The Natio

nal Lightening Detection Network used automatic sensors to watch for lightening in a sample of 16 randomly chosen square kilometers. Q1): What are the mean and standard deviation of the mean number of strikes per square kilometer
Mathematics
1 answer:
Alinara [238K]3 years ago
7 0

Answer:

5 ; 0.525

Step-by-step explanation:

Given that :

Population Mean number of strikes per square kilometer (μ) = 5

Population standard deviation of strikes per square kilometer (σ) = 2.1

Sample size (n) = 16

The sample mean number of strikes per square kilometer (m) = mean Number of strikes per square kilometer in the population.

μ = m ; 5 = 5

Hence,

Sample mean = 5

Standard deviation = standard error = σ/√n

Standard Error = 2.1 / √16

Standard Error = 2.1 / 4

Standard Error = 0.525

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Evaluate a(b - c ^ 2) * if * a = 2/3, b = 3/4, c = 1/2 A: 1/65 B: 1/3 C: 1/4 D: 2/3
DiKsa [7]

Answer:

If a+b+c=1,

a

2

+

b

2

+

c

2

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2

,

a

3

+

b

3

+

c

3

=

3

then find the value of

a

4

+

b

4

+

c

4

=

?

we know

2

(

a

b

+

b

c

+

c

a

)

=

(

a

+

b

+

c

)

2

−

(

a

2

+

b

2

+

c

2

)

⇒

2

(

a

b

+

b

c

+

c

a

)

=

1

2

−

2

=

−

1

⇒

a

b

+

b

c

+

c

a

=

−

1

2

given

a

3

+

b

3

+

c

3

=

3

⇒

a

3

+

b

3

+

c

3

−

3

a

b

c

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

a

b

−

b

c

−

c

a

)

+

3

a

b

c

=

3

⇒

(

a

+

b

+

c

)

(

a

2

+

b

2

+

c

2

−

(

a

b

+

b

c

+

c

a

)

+

3

a

b

c

=

3

⇒

(

1

×

(

2

−

(

−

1

2

)

+

3

a

b

c

)

)

=

3

⇒

(

2

+

1

2

)

+

3

a

b

c

=

3

⇒

3

a

b

c

=

3

−

5

2

=

1

2

⇒

a

b

c

=

1

6

Now

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

2

c

−

2

b

c

2

a

−

2

c

a

2

b

=

(

a

b

+

b

c

+

c

a

)

2

−

2

a

b

c

(

b

+

c

+

a

)

=

(

−

1

2

)

2

−

2

×

1

6

×

1

=

1

4

−

1

3

=

−

1

12

Now

a

4

+

b

4

+

c

4

=

(

a

2

+

b

2

+

c

2

)

2

−

2

(

a

2

b

2

+

b

2

c

2

+

c

2

a

2

)

=

2

2

−

2

×

(

−

1

12

)

=

4

+

1

6

=

4

1

6

Extension

a

5

+

b

5

+

c

5

=

(

a

3

+

b

3

+

c

3

)

(

a

2

+

b

2

+

c

2

)

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

c

2

)

]

=

3

⋅

2

−

[

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

]

Now

a

3

(

b

2

+

c

2

)

+

b

3

(

c

2

+

a

2

)

+

c

3

(

a

2

+

b

2

)

=

a

2

b

2

(

a

+

b

)

+

b

2

c

2

(

b

+

c

)

+

c

2

a

2

(

a

+

c

)

=

a

2

b

2

(

1

−

c

)

+

b

2

c

2

(

1

−

a

)

+

c

2

a

2

(

1

−

b

)

=

a

2

b

2

+

b

2

c

2

+

c

2

a

2

−

(

a

2

b

2

c

+

b

2

c

2

a

+

c

2

a

2

b

)

=

−

1

12

−

a

b

c

(

a

b

+

b

c

+

c

a

)

=

−

1

12

−

1

6

⋅

(

−

1

2

)

=

0

So

a

5

+

b

5

+

c

5

=

6

−

0

=

6

Step-by-step explanation:

8 0
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Answer:

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