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jarptica [38.1K]
3 years ago
10

8. Given the radius and area of the circle, calculate the angle at the centre of the circle.

Mathematics
1 answer:
pav-90 [236]3 years ago
5 0
The answer is c for this problem
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2w=56 what is in simplest form
Ivanshal [37]
2w= 56
/2 /2
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w= 28
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4 years ago
Can anyone please help with my last math problem pls all of my assignments gotta be turned in by tonight so I would really appre
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5 points is not enough

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Solve 2∕3 + 5∕6 and put answer in simplest form.<br> A. 3∕2<br> B. 2∕3<br> C. 9∕6<br> D. 7∕6
kaheart [24]
2/3 + 5/6
= 4/6 + 5/6
= 9/6
= 3/2

A. 3/2
6 0
4 years ago
For each given p, let Z have a binomial distribution with parameters p and N. Suppose that N is itself binomially distributed wi
8090 [49]

Explanation:

Lets interpret Z with M trials. First we have M trials, each trial can be a success or not. The number of success is called N. Each trial that is a success becomes a trial, and if it is a success it becomes a success for Z. Thus, in order for a trial to be successful, it needs first to be successful for the random variable N (and it is with probability q), and given that, it should be a success among the N trials of the original definition of Z (with probability p).

This gives us that each trial has probability pq of being successful. Note that this probability is pq independently of the results of the other trials, because the results of the trials of both N and the original definition of Z are independent. This shows us that Z is the total amount of success within M independent trials of an experiment with pq probability of success in each one. Therefore, Z has Binomial distribution with parameters pq and M.

7 0
3 years ago
Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,
sweet [91]

Answer:

Hence, the particular solution of the differential equation is y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x.

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

f'' = x - \frac{3}{2}

f' = \int {\left(x-\frac{3}{2}\right) } \, dx

f' = \int {x} \, dx -\frac{3}{2}\int \, dx

f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C, where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative (f'(4) = 1):

1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C

C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)

C = -1

The first derivative is y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1, and the particular solution is found by integrating one more time and using the initial condition (f(0) = 0):

y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1  \right)} \, dx

y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx

y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C

C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0

C = 0

Hence, the particular solution of the differential equation is y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x.

5 0
4 years ago
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