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2 years ago

What was the approximate percent decrease in the price of her groceries due to the coupons?

Mathematics

1 answer:

vovangra [49]2 years ago

3 0

Answer:

The approximate percent would be 25%

Step-by-step explanation:

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timothy bought 12 of his cars to the babysitters. these toy cars represent 30% of his toy car collection. what is the total numb

omeli [17]

If 30% of his collection is 12 cars, then we can use the fraction

30/12=100/x

x=40

7 0

3 years ago

How do you find the slope of y=-4/3x - 1?

Brut [27]

I can't think of what the form is called, but the slope is in that equation.

y=mx+b

m is the slope and in that equation, m=-4/3

8 0

3 years ago

Read 2 more answers

Which table shows a proportional relationship between x and y?

IrinaK [193]

The correct answer is the last set of numbers

6 0

2 years ago

Can I get some help please??

Fantom [35]

Answer:

$y = \frac{1}{4} x + 1$

Step-by-step explanation:

He got the slope wrong and the y-intercepz wrong

6 0

2 years ago

Read 2 more answers

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