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3 years ago

John flips 10 coins and lays them out in a straight line.

Mathematics

1 answer:

belka [17]3 years ago

3 0

For every coin, two options are possible. The formula for calculting all possibilities is $p^{t}$ where p is possibilities per time and t is times.

Therefore, there are $2^{10}$ which is 1024 combinations.

The formula for calculating probability is $\frac{w}{t}$ where w is the ways to achieve the desired outcome, and t is the total outcomes possible.

Therefore, there is a $\frac{1}{1024}$ chance of flipping ten tails.

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Nethan made 8 rose bouquets and 6 daffodil bouquets. Nethan only has enough flowers to make at most 20 rose or daffodil bouquets

Solnce55 [7]

Answer:

Nethan made 8 rose bouquets and 6 daffodil bouquets.

Nethan only has enough flowers to make at most 20 rose or daffodil bouquets total.

Let x represents the number of more rose bouquets.

Let y represents the number of more daffodil bouquets.

The maximum bouquets that can be made = 20

And the number of bouquets made are = 14

So, additional bouquets are = $20-14=6$

In equation form we can show like :

$x +y= 6$

or $y=6 -x$

or $x=6- y$

Putting values of x as 0,2,4 and 6 we get:

When x = 0 then y = 6

When x = 2 then y = 4

When x = 4 then y = 2

When x = 6 then y = 0

You can see the graph attached.

3 0

3 years ago

Thomas spent $3.30 for 3 ice cream cones. Dan spent 2.50 for 2 ice cream cones. Who spent less per ice cream cone

dsp73

Answer:

Thomas spent less per ice cream cone.

Step-by-step explanation:

Thomas spent $1.10 per ice cream cone.

Dan spent $1.25 per ice cream cone.

8 0

3 years ago

Please help This homework is NOT for a grade so please please PLEASE help!

Marta_Voda [28]

Answer:

Hey if its not for a grade why do it??

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Let f be the function defined by f(x) = e^(x) cos x.

Pavel [41]

(a)

The average rate of change of f on the interval 0 ≤ x ≤ π is

$\displaystyle
f_{avg\Delta} = \frac{f(\pi) - f(0)}{\pi - 0} =\frac{-e^\pi-1}{\pi}$

____________

(b)

$f(x) = e^{x} cos x \implies f'(x) = e^x \cos(x) - e^x \sin(x) \implies \\ \\
f'\left(\frac{3\pi}{2} \right) = e^{3\pi/2} \cos(3\pi/2) - e^{3\pi/2} \sin(3\pi/2) \\ \\
f'\left(\frac{3\pi}{2} \right) = 0 - e^{3\pi/2} (-1) = e^{3\pi/2}$

The slope of the tangent line is $e^{3\pi/2}$ .

____________

(c)

The absolute minimum value of f occurs at a critical point where f'(x) = 0 or at endpoints.

Solving f'(x) = 0

$f'(x) = e^x \cos(x) - e^x \sin(x) \\ \\
0 = e^x \big( \cos(x) - \sin(x)\big)$

Use zero factor property to solve.

$e^x \ \textgreater \ 0\forall x \in \mathbb{R}$ so that factor will not generate solutions.
Set cos(x) - sin(x) = 0

$\cos (x) - \sin (x) = 0 \\
\cos(x) = \sin(x)$

cos(x) = 0 when x = π/2, 3π/2, but x = π/2. 3π/2 are not solutions to the equation. Therefore, we are justified in dividing both sides by cos(x) to make tan(x):

$\displaystyle\cos(x) = \sin(x) \implies 0 = \frac{\sin (x)}{\cos(x)} \implies 0 = \tan(x) \implies \\ \\
x = \pi/4,\ 5\pi/4\ \forall\ x \in [0, 2\pi]$

We check the values of f at the end points and these two critical numbers.

$f(0) = e^1 \cos(0) = 1$

$\displaystyle f(\pi/4) = e^{\pi/4} \cos(\pi/4) = e^{\pi/4} \frac{\sqrt{2}}{2}$

$\displaystyle f(5\pi/4) = e^{5\pi/4} \cos(5\pi/4) = e^{5\pi/4} \frac{-\sqrt{2}}{2} = -e^{\pi/4} \frac{\sqrt{2}}{2}$

$f(2\pi) = e^{2\pi} \cos(2\pi) = e^{2\pi}$

There is only one negative number.
The absolute minimum value of f <span>on the interval 0 ≤ x ≤ 2π is
$-e^{5\pi/4} \sqrt{2}/2$

____________

(d)

The function f is a continuous function as it is a product of two continuous functions. Therefore, $\lim_{x \to \pi/2} f(x) = f(\pi/2) = e^{\pi/2} \cos(\pi/2) = 0$

g is a differentiable function; therefore, it is a continuous function, which tells us $\lim_{x \to \pi/2} g(x) = g(\pi/2) = 0$ .

When we observe the limit $\displaystyle \lim_{x \to \pi/2} \frac{f(x)}{g(x)}$ , the numerator and denominator both approach zero. Thus we use L'Hospital's rule to evaluate the limit.

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \lim_{x \to \pi/2} \frac{f'(x)}{g'(x)} = \frac{f'(\pi/2)}{g'(\pi/2)}$

$f'(\pi/2) = e^{\pi/2} \big( \cos(\pi/2) - \sin(\pi/2)\big) = -e^{\pi/2} \\ \\
g'(\pi/2) = 2$

thus

$\displaystyle\lim_{x \to \pi/2} \frac{f(x)}{g(x)} = \frac{-e^{\pi/2}}{2}$ </span>

3 0

3 years ago

5b-b-1 > or equal too-11

vova2212 [387]

Answer:

-2.5

Step-by-step explanation:

5b - b - 1 ≥ -11

4b - 1 ≥ -11

+1 +1

4b ≥ -10

4/4 -10/4

b ≥ -2.5

4 0

3 years ago

Read 2 more answers