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ivann1987 [24]
3 years ago
9

If there is a 30% chance of sun tomorrow and a 40% chance of wind and no sun, what is the probability that it is windy, given th

at it is not sunny? Round your answer to the nearest percent.
A. 57%
B. 29%
C. 22%
D. 44%
Mathematics
2 answers:
stira [4]3 years ago
6 0
A, 57%, because 0.40 divided by all possible outcomes, which is 0.70 because we know it will not be sunny tomorrow, is 0.57.
Karo-lina-s [1.5K]3 years ago
3 0
The answer is 57% APEX
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Consider the solution to the linear equation. 5 (x + 6) = 50. 5 x + 30 = 50. 5 x = 20. x = 4. Which describes the inverse operat
dem82 [27]

(trying to isolate/get x by itself in the equation)

5(x + 6) = 50      Distributive property   [distribute 5 into (x + 6)]

5x + 30 = 50    Subtraction    [subtract 30 on both sides of the equation]

5x = 20       Division     [divide 5 on both sides]

x = 4

Subtraction then division, the 2nd option

7 0
3 years ago
Read 2 more answers
Pls help with this if it is not to hard for you to do ​
SashulF [63]

Answer:

12x-3y

Step-by-step explanation:

10x3 + x + X

Group like terms

10x + x +X-3

Add similar elements: 10x + x + x = 12x

= 12x - 3y

5 0
2 years ago
Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.
Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Step-by-step explanation:

GIven that:

f(x) = 5e^{-x^2} cos (4x)

The Maclaurin series of cos x can be expressed as :

\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+...  \ \ \ (1)}

\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0}  \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!}  -\dfrac{x^6}{3!}+... \ \ \  (2)}

From equation(1), substituting x with (4x), Then:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}

The first three terms of cos (4x) is:

\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}

\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}

\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}

Multiplying equation (2) with (3); we have :

\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }

\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }

\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

Finally , multiplying 5 with \mathtt{ e^{-x^2} cos (4x) } ; we have:

The first three nonzero terms in the Maclaurin series is

\mathbf{ 5e^{-x^2} cos (4x)  }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }

7 0
3 years ago
Suppose you went to a carnival. The price to get in was $4, and you paid $0.50 to get on each ride. If you went to the carnival
Strike441 [17]
The answer is b $7.50
4 + 7(0.50)
4+3.5
7.50
6 0
3 years ago
Read 2 more answers
I need sum help........
zimovet [89]

Answer:

hey! id love to help but its not loading. do you have a link by any chance?

Step-by-step explanation:

7 0
3 years ago
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