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3 years ago

Which three criteria do binomial experiments meet?

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

3 0

Answer:

Each of the n tests must have two success p or failure q results, they are mutually exclusive. The probability of success p of an event is fixed or constant, likewise for the probability of failure q. The tests are independent.

Step-by-step explanation:

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Please help I'll give brainliest

SIZIF [17.4K]

Answer:

y + 1 = 4(x - 2)

y = 4x + 11

8x - 2y = 6

Step-by-step explanation:

Following equations are parallel to the graph 4x - y = 6, because their slopes are equal (4).

y + 1 = 4(x - 2)

y = 4x + 11

8x - 2y = 6

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3 years ago

MIDDLE SCHOOL MATH BRAINLEIST AND 5 STARS AS SOON AS YOU ANSWER!!!!!!!! PLEASE HELP AND THANKS SO MUCH IM SUPER GRATEFUL!!!!!!!!

gladu [14]

Answer:

Bottom prism: 126*11=1386

Top Prism: 2*45=90

Total volume: 1386+90= 1746 cubic cm

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3 years ago

Read 2 more answers

Which of the following options have the same value as 2% of 90? Choose 2 answers: А 0.2.90 B 0.02.90 200.90 2.90 E 2 . 90 100

Wewaii [24]

Answer:

0.02.90 and 2.90

Step-by-step explanation:

2% of 90 equal to both 0.02.90 because are worth same.

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3 years ago

A company uses tow vans to transport workers from a free parking lot to the workplace between 7:00 and 9:00 a.m. one van has 6 m

sladkih [1.3K]

The large van has 23 seats and the small van has 17 seats

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3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

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3 years ago