All categories

3 years ago

Fortyeight multiplied by twevle

Mathematics

2 answers:

tresset_1 [31]3 years ago

8 0

Hello there!

Answer: =576

Word question spelled right: → Forty-eight times twelve.

48*12=96+480=576

Explanation: Multiply by the numbers. Product it's going to be multiply by the numbers, fractions, decimals, factors, and multiples. This gives us 96+480=576 is the final answer.

Hope this helps!

Thanks!

Have a great day!

expeople1 [14]3 years ago

5 0

<h2>48 x 12 =576 that is the correct answer </h2>

You might be interested in

What number comes next? 2,7,8,3,12,9

s2008m [1.1K]

4,17,9. That is the answer

4 0

4 years ago

1. 125 decreased by a number

Volgvan

Answer:

125-x

Step-by-step explanation:

i don't know what exactly you're aiming for, but if you want to write this as an equation, you would put it as 125-x. since we don't know what the number is, we would put a variable that we can sub in any number we want.

hope this made sense!

3 0

3 years ago

Read 2 more answers

Consider the simple linear regression model Yi=β0+β1xi+ϵi, where ϵi's are independent N(0,σ2) random variables. Therefore, Yi is

Virty [35]

Answer:

See proof below.

Step-by-step explanation:

If we assume the following linear model:

$y = \beta_o + \beta_1 X +\epsilon$

And if we have n sets of paired observations $(x_i, y_i) , i =1,2,...,n$ the model can be written like this:

$y_i = \beta_o +\beta_1 x_i + \epsilon_i , i =1,2,...,n$

And using the least squares procedure gives to us the following least squares estimates $b_o$ for $\beta_o$ and $b_1$ for $\beta_1$ :

$b_o = \bar y - b_1 \bar x$

$b_1 = \frac{s_{xy}}{s_xx}$

Where:

$s_{xy} =\sum_{i=1}^n (x_i -\bar x) (y-\bar y)$

$s_{xx} =\sum_{i=1}^n (x_i -\bar x)^2$

Then $\beta_1$ is a random variable and the estimated value is $b_1$ . We can express this estimator like this:

$b_1 = \sum_{i=1}^n a_i y_i$

Where $a_i =\frac{(x_i -\bar x)}{s_{xx}}$ and if we see careful we notice that $\sum_{i=1}^n a_i =0$ and $\sum_{i=1}^n a_i x_i =1$

So then when we find the expected value we got:

$E(b_1) = \sum_{i=1}^n a_i E(y_i)$

$E(b_1) = \sum_{i=1}^n a_i (\beta_o +\beta_1 x_i)$

$E(b_1) = \sum_{i=1}^n a_i \beta_o + \beta_1 a_i x_i$

$E(b_1) = \beta_1 \sum_{i=1}^n a_i x_i = \beta_1$

And as we can see $b_1$ is an unbiased estimator for $\beta_1$

In order to find the variance for the estimator $b_1$ we have this:

$Var(b_1) = \sum_{i=1}^n a_i^2 Var(y_i) +\sum_i \sum_{j \neq i} a_i a_j Cov (y_i, y_j)$

And we can assume that $Cov(y_i,y_j) =0$ since the observations are assumed independent, then we have this:

$Var (b_1) =\sigma^2 \frac{\sum_{i=1}^n (x_i -\bar x)^2}{s^2_{xx}}$

And if we simplify we got:

$Var(b_1) = \frac{\sigma^2 s_{xx}}{s^2_{xx}} = \frac{\sigma^2}{s_{xx}}$

And with this we complete the proof required.

8 0

4 years ago

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro

Gnom [1K]

Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.