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3 years ago

Alton is going to flip a fair coin 350 times.

Mathematics

1 answer:

Arlecino [84]3 years ago

7 0

Answer:

175 times

Step-by-step explanation:

To do this, we can find the probability of the coin landing tails up.

There is 1 desired outcome: Landing tails up.

There are 2 total outcomes: Landing tails up or heads up.

The probability is 1/2 for the coin to land tails up every time.

1/2*350=175

The coin will land tails up 175 times.

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A man has 14 coins in his pocket, all of which are dimes and quarters. If the total value of his

S_A_V [24]

Answer:

9 quarters 5 dimes!

Step-by-step explanation:

9x0.25=2.25

5x0.10=0.50

2.25+0.50=2.75

6 0

2 years ago

Five tickets to a baseball game cost 60 dollars how much does 2 tickets cost

LUCKY_DIMON [66]

Five divided by sixty is twelve, and twelve times two is twenty-four. So your answer will be twenty-four (24).

4 0

3 years ago

Read 2 more answers

What is The cube of the product of 3 and x

Anni [7]

Answer: 9

Step-by-step explanation:

5 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%20lnx%20%7D%20" id="TexFormula1" title=" \sqrt[4]{ lnx } " alt=" \sqrt[4]

Kamila [148]

Use the power rule for differentiation:

$\dfrac{\text{d}}{\text{d}x} (f(x))^k = k(f(x))^{k-1}f'(x)$

You can use this formula if you remember that a root is just a rational exponential:

$\sqrt[4]\ln(x) = (\ln(x))^{\frac{1}{4}}$

So, remembering that the derivative of the logarithm is 1/x, you have

$\dfrac{\text{d}}{\text{d}x} (\ln(x))^{\frac{1}{4}} = \frac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x}$

Which you can rewrite as

$\dfrac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x} =\dfrac{1}{4}(\ln(x))^{\frac{-3}{4}}\dfrac{1}{x} =\dfrac{1}{4}\dfrac{1}{\sqrt[4]{\ln(x))^3}}\dfrac{1}{x} = \dfrac{1}{4x\sqrt[4]{\ln(x))^3}}$

3 0

2 years ago

Graph triangle RST with vertices R(3, 7), S(-5, -2), and T(3, -5) and its image after a reflection over x = -3.

kirill115 [55]

Given:

The vertices of a triangle are R(3, 7), S(-5, -2), and T(3, -5).

To find:

The vertices of the triangle after a reflection over x = -3 and plot the triangle and its image on the graph.

Solution:

If a figure reflected across the line x=a, then

$(x,y)\to (-(x-a)+a,y)$

$(x,y)\to (-x+a+a,y)$

$(x,y)\to (2a-x,y)$

The triangle after a reflection over x = -3. So, the rule of reflection is

$(x,y)\to (2(-3)-x,y)$

$(x,y)\to (-6-x,y)$

The vertices of triangle after reflection are

$R(3,7)\to R'(-6-3,7)$

$R(3,7)\to R'(-9,7)$

Similarly,

$S(-5,-2)\to S'(-6-(-5),-2)$

$S(-5,-2)\to S'(-6+5,-2)$

$S(-5,-2)\to S'(-1,-2)$

And,

$T(3,-5)\to T'(-6-3,-5)$

$T(3,-5)\to T'(-9,-5)$

Therefore, the vertices of triangle after reflection over x=-3 are R'(-9,7), S'(-1,-2) and T'(-3,-5).

3 0

2 years ago