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3 years ago

I need help this is dude tomorrow

Mathematics

1 answer:

lara31 [8.8K]3 years ago

8 0

Don’t listen to them it’s a scam

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A coin is flipped until 3 heads in succession occur. list only those elements of the sample space that require 6 or less tosses.

Mashutka [201]

we are given

A coin is flipped until 3 heads in succession occur

so, firstly, we will find sample space

S={HHH , THHH , HTHHH, TTHHH , TTTHHH , HTTHHH , THTHHH , HHTHHHH, ......}

now, we are given that

list only those elements of the sample space that require 6 or less tosses

so, we can see that sample space

S={HHH , THHH , HTHHH, TTHHH , TTTHHH , HTTHHH , THTHHH , HHTHHHH, ......}

There are infinite such possibilities

so, there are infinite number of elements in space

and we know that

discrete sample space will always have finite elements

but we have infinite number of sample space elements here

so, this is not discrete sample space...........Answer

6 0

3 years ago

Need help pls help me

Maurinko [17]

Answer:

A) 7, 2

B) 1, 5

C) 6, 3

6 0

3 years ago

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Solve for x. 79 (8x - 4) 3 -(3x + 17)

Liono4ka [1.6K]

Answer:

-2.4

Step-by-step explanation:

maybe... i am 99.99% sure

6 0

2 years ago

Find the roots of the equation x ^ 2 + 3x-8 ^ -14 = 0 with three precision digits

scoray [572]

Answer:

Step-by-step explanation:

Given quadratic equation:

$x^{2} + 3x - 8^{- 14} = 0$

The solution of the given quadratic eqn is given by using Sri Dharacharya formula:

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

The above solution is for the quadratic equation of the form:

$ax^{2} + bx + c = 0$

$x_{1, 1'} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$

From the given eqn

a = 1

b = 3

c = $- 8^{- 14}$

Now, using the above values in the formula mentioned above:

$x_{1, 1'} = \frac{- 3 \pm \sqrt{3^{2} - 4(1)(- 8^{- 14})}}{2(1)}$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})})$

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(1)(- 8^{- 14})} - 3)$

Now, Rationalizing the above eqn:

$x_{1, 1'} = \frac{1}{2} (\pm \sqrt{9 - 4(- 8^{- 14})} - 3)\times (\frac{\sqrt{9 - 4(- 8^{- 14})} + 3}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

$x_{1, 1'} = \frac{1}{2}.\frac{(\pm {9 - 4(- 8^{- 14})^{2}} - 3^{2})}{\sqrt{9 - 4(- 8^{- 14})} + 3}$

Solving the above eqn:

$x_{1, 1'} = \frac{2\times 8^{- 14}}{\sqrt{9 + 4\times 8^{-14}} + 3}$

Solving with the help of caculator:

$x_{1, 1'} = \frac{2\times 2.27\times 10^{- 14}}{\sqrt{9 + 42.27\times 10^{- 14}} + 3}$

The precise value upto three decimal places comes out to be:

$x_{1, 1'} = 0.758\times 10^{- 14}$

5 0

3 years ago

Jim hit the ball 9 out of 15 times. Find the percent Jim hit the ball.

lapo4ka [179]

Answer:

9/15 = 0.6 = 60%

Step-by-step explanation:

5 0

3 years ago

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