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3 years ago

7

If you flip a coin 4 times, what is the probability of flipping heads 4 times?

1 answer:

finlep [7]3 years ago

8 0

1/2 chance per flip
1/2 times 4 is 1/8
its 1/8 chance or 12.5%

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Find the sum of the series and show your process for finding each term leading to the final sum.

Crazy boy [7]

ANSWER

$\sum_{k=3}^5( - 2k + 5) = - 9$

EXPLANATION

The given series is

$\sum_{k=3}^5( - 2k + 5)$

This series is finite.

The expanded form is

$\sum_{k=3}^5( - 2k + 5) = ( - 2 \times 3 + 5) + ( - 2 \times 4 + 5) + ( - 2 \times 5 + 5)$

This simplifies to

$\sum_{k=3}^5( - 2k + 5) = ( - 6 + 5) + ( - 8+ 5) + ( - 10+ 5)$

$\sum_{k=3}^5( - 2k + 5) = ( - 1) + ( - 3) + ( - 5)$

$\sum_{k=3}^5( - 2k + 5) = - 9$

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2 years ago

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

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2 years ago

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Svet_ta [14]

Answer:

P = 24

Step-by-step explanation:

P = 2L + 2W ( substitute L = 4, W = 8 into the formula )

P = 2(4) + 2(8) = 8 + 16 = 24

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kotegsom [21]

Answer:

1. x =−3, -4, -5

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