For the smaller triangle ;
15^2 + b^2 = 17^2
b = 8

To check the work
15^2 + 8^2 = 289
17^2 = 289

(This is if the bottom number on the smaller triangle says 15, its a bit hard to read..)

6 0

3 years ago

To save money, you put $200 in your bank account each week. After saving for 4 weeks, you have $1,700 dollars in your account. W

kati45 [8]

it is c y−1,700=200(x−4)

8 0

2 years ago

How do I solve this one

just olya [345]

6/7Y - 18/28 =-15/14
First, let's get a common denominator
24/28Y - 18/28 =-30/28
Move your terms to isolate Y
24/28Y=-12/28
Multiply by 28 on both sides
24/Y=-12
Times both sides by Y, then divide by -12
24/-12=Y
Simplify
-2=Y

I hope this Helps!

6 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

3 years ago