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3 years ago

In a binomial experiment, the outcome of each trial will depend on the ones before it.

Mathematics

1 answer:

ra1l [238]3 years ago

7 0

The answer would be false

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Help help help math math

Rina8888 [55]

Answer:

100 cm³

Step-by-step explanation:

If you use 3 for π, then the factors of 3 cancel in the formula for volume, and you have ...

V = r²h

The radius is half the diameter, so is (10 cm)/2 = 5 cm. Filling in the given values, you find the volume to be ...

V = (5 cm)²(4 cm) = 100 cm³

7 0

3 years ago

:P ??????<br><br><br><br><br><br> “ItS ToO ShOrT”

Rus_ich [418]

Answer:

The answer is >.

Step-by-step explanation:

78 19/88 is bigger than 74/93 because 74/93 has way less whole numbers on its side.

4 0

3 years ago

Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$

8 0

1 year ago

A university's freshman class has 6200 students. 4836 of those students are majoring

WINSTONCH [101]

According to the solving 78% of the freshman class are Economics majors.

<h3>What does a percentage mean?</h3>

percentage, a relative figure signifying hundredths of any amount. Since one percent (symbolized as 1%) is equal to one hundredth of something, 100 percent stands for everything, and 200 percent refers to twice the amount specified.

<h3>Why is it called percent?</h3>

The word "percent" comes from the Latin adverb per centum, which means "by the hundred." In the sixteenth century, the Latin expression made its way into English. Later, it was shortened to % with a final period.

<h3>According to the given data:</h3>

It is given that the total number of students in freshman class = 6200

Total number of students majoring in Economics = 4836

Here, we are asked to find the percentage of the freshman class who are majoring in Economics.

Percent of students majoring Economics = (Number of students majoring/ Economics Total number of students) x 100

= $\frac{4836}{6200}$ x 100

= 78%

According to the solving 78% of the freshman class are Economics majors.

To know more about Percentage visit:

brainly.com/question/14979505

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4 0

1 year ago

Hi people I need help on this question can y’all help me pls!!!!!! Thanks

skad [1K]

I’m pretty sure it’s 43.96 I may be wrong though!

6 0

3 years ago