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2 years ago

Problem 2 Find the mBC.

Mathematics

1 answer:

MatroZZZ [7]2 years ago

7 0

Answer:

38º

Step-by-step explanation:

<BAC is a central angle. The corresponding arc, BC will be equal to the measure of the central angle.

So mBC = 38º

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What is the sum of the first 11 terms of the arithmetic series in which a1= -12 and d=5?

SVEN [57.7K]

Hello :
the terme general is : an = a1 +(n-1)d
a11 = a1+(11-1)×5
a11 = -12+50
a11 = 38
<span>the sum of the first 11 terms is : S11 = (11/2)(a1+a11)
</span>S11 = (11/2)(-12+38)
S11 = 143

3 0

2 years ago

Subtract (5x^2-5) from the sum (x^2-9x+2) and (4x^2-5x+1)

Ivan

Assignment: $\bold{Solve \ Equation: \ \left(x^2-9x+2\right)+\left(4x^2-5x+1\right)-\left(5x^2-5\right)}$

<><><><><><><>

Answer: $\boxed{\bold{-14x+8}}$

<><><><><><><>

Explanation: $\downarrow\downarrow\downarrow$

<><><><><><><>

[ Step One ] Remove Parenthesis: (a) = a

$\bold{x^2-9x+2+4x^2-5x+1-\left(5x^2-5\right)}$

[ Step Two ] Simplify $\bold{-\left(5x^2-5\right)}$

$\bold{-5x^2+5}$

[ Step Three ] Rewrite Equation

$\bold{x^2-9x+2+4x^2-5x+1-5x^2+5}$

[ Step Four ] Simplify $\bold{x^2-9x+2+4x^2-5x+1-5x^2+5}$

$\bold{-14x+8}$

<><><><><><><>

$\bold{\rightarrow Mordancy \leftarrow}$

3 0

2 years ago

The sum of nine and three times a number is 57 find the number and write an equation

Temka [501]

Let "n" be a number

9 + 3n = 57
3n = 57 - 9
3n = 48
n = 48/3
n = 16

7 0

2 years ago

Let X1, X2, ... , Xn be a random sample from N(μ, σ2), where the mean θ = μ is such that −[infinity] < θ < [infinity] and

Sliva [168]

Answer:

$l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)$

And then the maximum occurs when $l'(\theta) = 0$ , and that is only satisfied if and only if:

$\hat \theta = \bar X$

Step-by-step explanation:

For this case we have a random sample $X_1 ,X_2,...,X_n$ where $X_i \sim N(\mu=\theta, \sigma)$ where $\sigma$ is fixed. And we want to show that the maximum likehood estimator for $\theta = \bar X$ .

The first step is obtain the probability distribution function for the random variable X. For this case each $X_i , i=1,...n$ have the following density function:

$f(x_i | \theta,\sigma^2) = \frac{1}{\sqrt{2\pi}\sigma} exp^{-\frac{(x-\theta)^2}{2\sigma^2}} , -\infty \leq x \leq \infty$

The likehood function is given by:

$L(\theta) = \prod_{i=1}^n f(x_i)$

Assuming independence between the random sample, and replacing the density function we have this:

$L(\theta) = (\frac{1}{\sqrt{2\pi \sigma^2}})^n exp (-\frac{1}{2\sigma^2} \sum_{i=1}^n (X_i-\theta)^2)$

Taking the natural log on btoh sides we got:

$l(\theta) = -\frac{n}{2} ln(\sqrt{2\pi\sigma^2}) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i -\theta)^2$

Now if we take the derivate respect $\theta$ we will see this:

$l'(\theta) = \frac{1}{\sigma^2} \sum_{i=1}^n (X_i -\theta)$

And then the maximum occurs when $l'(\theta) = 0$ , and that is only satisfied if and only if:

$\hat \theta = \bar X$

6 0

2 years ago

I need help on this for math class dont mind how i answered a

Scrat [10]

Step-by-step explanation:

I think its c cus its more relatable .

7 0

2 years ago

Read 2 more answers