All categories

3 years ago

33.3% coverted into a decimle and fraction

2 answers:

5 0

-- To change a percent into a decimal, move
the decimal point two places this way <== .

   33.3% = 0.333 .

-- To change a percent into a fraction, remember
that the symbol ' % ' means ' /100 ' .

   33.3 % = 33.3/100 .

That fraction looks ugly, because it has a decimal in the numerator.
That's easy to fix. Multiply top and bottom by 10, and you have

   333/1000 .

Julli [10]3 years ago

5 0

<span>You get 333/1000 because you move two decimal places to the left in 33.3 % and then convert it to fraction.</span>

You might be interested in

HEEELP MEEEE FIND THE CIRCUMFERENCE

Alex Ar [27]

Answer: 56.52

Step-by-step explanation:

4 0

3 years ago

4 × (28 ÷ 7)2 = <br> HURRYYYYY

umka21 [38]

The answer is 32

hope this helps! :)

8 0

3 years ago

Imagine a couple who is ready to start a family. They plan to have exactly four children. Assuming no multiple births (twins, tr

Lostsunrise [7]

Answer:

4 different ways

Step-by-step explanation:

Total number of children = 4

Distribution of the 4 children :

Number of boys = 3 ; Number of girls = 1

Boy = B ; Girl = G

Possible combinations :

BBBG ; GBBB ; BBGB ; BGBB

From the pascal triangle number of e; number of outcomes = 2

Having exactly 3 boys and 1 girl

Hence, of any of the 4 four total children, 3 must be boys and 1 girl ;

7 0

3 years ago

Can you help me please?

KIM [24]

Answer:

option c.

n=12

.................................

7 0

3 years ago

Let X1,X2......X7 denote a random sample from a population having mean μ and variance σ. Consider the following estimators of μ:

Viefleur [7K]

Answer:

a) In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

b) For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

Step-by-step explanation:

For this case we assume that we have a random sample given by: $X_1, X_2,....,X_7$ and each $X_i \sim N (\mu, \sigma)$

Part a

In order to check if an estimator is unbiased we need to check this condition:

$E(\theta) = \mu$

And we can find the expected value of each estimator like this:

$E(\theta_1 ) = \frac{1}{7} E(X_1 +X_2 +... +X_7) = \frac{1}{7} [E(X_1) +E(X_2) +....+E(X_7)]= \frac{1}{7} 7\mu= \mu$

So then we conclude that $\theta_1$ is unbiased.

For the second estimator we have this:

$E(\theta_2) = \frac{1}{2} [2E(X_1) -E(X_3) +E(X_5)]=\frac{1}{2} [2\mu -\mu +\mu] = \frac{1}{2} [2\mu]= \mu$

And then we conclude that $\theta_2$ is unbiaed too.

Part b

For this case first we need to find the variance of each estimator:

$Var(\theta_1) = \frac{1}{49} (Var(X_1) +...+Var(X_7))= \frac{1}{49} (7\sigma^2) = \frac{\sigma^2}{7}$

And for the second estimator we have this:

$Var(\theta_2) = \frac{1}{4} (4\sigma^2 -\sigma^2 +\sigma^2)= \frac{1}{4} (4\sigma^2)= \sigma^2$

And the relative efficiency is given by:

$RE= \frac{Var(\theta_1)}{Var(\theta_2)}=\frac{\frac{\sigma^2}{7}}{\sigma^2}= \frac{1}{7}$

5 0

4 years ago