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Nimfa-mama [501]

3 years ago

8

Amanda purchased a prepaid phone card for $20. Long distance calls cost 11 cents a minute using this card. Amanda used her card

only
once to make a long distance call. If the remaining credit on her card is $16.81, how many minutes did her call last?
minutes

1 answer:

lutik1710 [3]3 years ago

5 0

Answer:

29 minutes?

i divided 3.19 by 11 (20 - 16.81 is 3.19) and i got 29 so i think it is that amount of minutes

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There are 210 songs in Jane's notebook, while she has 150 songs in her mobile phone. How many songs she must transfer from the n

zhannawk [14.2K]

Answer:

The answer is 90 songs.

Step-by-step explanation:

Let x be the number she has to transfer, then:

x + 150 =2(210 - x)

x + 150 = 420 - 2x

3x = 420 - 150

3x = 270

x = 90

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The graph shows the growth of a tree, with x representing the number of years since it was planted, and y representing the tree’

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The tree was 40 in tall when planted

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Step-by-step explanation:

original(slope intercept form) : y = 10x + 40

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4 years ago

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Bit transmission errors between computers sometimes occur, where one computer sends a 0 but the other computer receives a 1 (or

lana66690 [7]

Answer:

(a) Probability that a triplet is decoded incorrectly by the receiving computer. = 0.010

(b)

(1 – p) = 0.010

(c)

E(x) = 25000 x 0.010

= 259.2

Explanation has given below.

Step-by-step explanation:

Solution:

(a) Probability that a triplet is decoded.

2 out of three

P = 0.94, n = 3

m= no of correct bits

m bit (3, 0.94)

At p(m≤1) = B (1; 3, 0.94)

= 0.010

(b) Using your answer to part (a),

(1 – p) = 0.010

Error for 1 bit transmission error.

(c) How does your answer to part (a) change if each bit is repeated five times (instead of three?

P( m ≤ 2 )

L = Bit (5, 0.94)

= B (2; 5, 0.94)

= 0.002

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Given:

h = 25000

Bits are switched during transmission between two computers = 6% = 0.06

m = Bit (25000, 0.06)

E(m) = np

= 25000 x 0.06

= 1500

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E(m) = 25000 x 0.010

= 259.2

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3 years ago

Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{12},0)(± 12 ,0)left parenthesis, plus min

lora16 [44]

Answer:

$\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}$

Step-by-step explanation:

Given that :

the foci of the ellipse is (±√12,0) and C0-vertices are (0,±√37)

The foci are (-C,0) and (C ,0)

the focus has x-coordinates so the focus is lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so C0-vertices are (0,±√37)

The given focus C = ae = √12

Given co-vertices ( minor axis) (0,±b) = (0,±√37)

b= √37

We can therefore express the relation between the focus and semi major axes and semi minor axes as:

$\mathbf{c^2 = a^2 - b^2 } \\ \\ \mathbf{a^2 = c^2 + b^2 } \\ \\ \mathbf{c^2 = ( \sqrt12)^2 - (\sqrt 37)^2 } \\ \\ \mathbf{c^2 = 49 } \\ \\ \mathbf{c = \sqrt{49 }}$

The equation of ellipse formula is:

$\dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

and we know that $\mathbf{a=\sqrt{49} \ \ and \ \ b=\sqrt{37}}$

Thus ; the equation of the ellipse at the origin is

$\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}$

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