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dexar [7]
3 years ago
6

Variance 0.7775

Mathematics
2 answers:
nadezda [96]3 years ago
6 0

Answer:

0.88175960442

Step-by-step explanation:

The square root of 0.7775 is 0.88175960442

rewona [7]3 years ago
4 0

Answer:

0.88

Step-by-step explanation:

To find the standard deviation square root the variance,

SD=\sqrt{0.7775}\\ SD= 0.88

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it's because we should devid eit for calculation for minute

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Concentric circles are circles with the same center but different radii. Which equations represent concentric circles along with
Harrizon [31]
<span>(x – 8)² + (y – 9)² = 3
(x – 8)² + (y – 9)² = 14
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Write an equation for each context. Draw a model first if necessary, and make sure you define your variable and what it represen
Bad White [126]

Answer:

(a)$2248.8

(b) 14, 28, 16 and 7 candies

(c) $79

Step-by-step explanation:

(a) the model can be represented by the nth term of an arithmetic progression

L = a + (n-1)d

L= the amount in his account after 6 month; $12484

a= the amount in his account six month before; $1240

n = 6, being the sixth month of his deposit

d = the amount deposited monthly

12484 = 1240 + 5d:

12484-1240 = 5d

5d= 11244

d = 2248.8

(b) let the 1st box contain = x

2nd = 2x; 3rd = x + 2, 4th = 1/4(2x)

Hence; x + 2x + x +2 + 1/4(2x) = 65

4x + x/2 = 65-2

9x/2 = 63 ; 9x = 126. ; x = 14

(c) Each player gets a racket (r) and a can (c) of balls

The 5 players get in all 5 racket + 5 cans since Simon is a member

Therefore, 5r + 5c = 415

But each can cost $4,

5r + 5* 4 = 415; 6r = 415-20

5r = 395 ; r = 395/5

r = $ 79

The racket cost $79 each.

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Step-by-step explanation:

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For what values of b are the vectors [−18, b, 9] and [b, b2, b] orthogonal? (Enter your answers as a comma-separated list. If an
bearhunter [10]

Answer:

Therefore the given vectors are orthogonal for b = 0,±3.

Step-by-step explanation:

If  \vec a and  \vec b are two vectors orthogonal, then the dot product of \vec a and \vec b will be zero.

i.e \vec a. \vec b =0

If  \vec a = x_1\hat i+y_1\hat j +z_1\hat k  and \vec b = x_2\hat i+y_2\hat j +z_2\hat k

\vec a. \vec b=( x_1\hat i+y_1\hat j +z_1\hat k).( x_2\hat i+y_2\hat j +z_2\hat k)

     =x_1 x_2+y_1y_2+z_1z_2

Given two vectors are (-18,b,9) and (b,b²,b)

Let

\vec P= -18 \hat i+b\hat j +9 \hat k

and

\vec Q = b \hat i+b^2 \hat j +b\hat k

Therefore,

\vec P.\vec Q

=( -18 \hat i+b\hat j +9 \hat k).( b \hat i+b^2 \hat j +b\hat k)

=(-18).b+b.b²+9.b

= -18b+b³+9b

= b³-9b

Since \vec P and \vec Q are orthogonal. Then \vec P.\vec Q = 0.

Therefore,

b³-9b= 0

⇒b(b²-9)=0

⇒b =0 or b²=9

⇒b=0 or b =±3

Therefore the given vectors are orthogonal for b = 0,±3.

8 0
3 years ago
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