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3 years ago

8

Craig likes to collect records. last year he had 10 records in his collection. now he has 13 records. what is the percent increa

se of his collection?

2 answers:

Mrac [35]3 years ago

8 0

Answer:

30%

Step-by-step explanation:

lol

Eva8 [605]3 years ago

5 0

Answer:

The answer to this problem would be 30%

Step-by-step explanation:

<em> I had this exact problem, and I got it correct......</em>

You might be interested in

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

3 years ago

WILL GIVE BRAINLIEST

Maksim231197 [3]

Just change -1/5 to a decimal and then plot them in the graph

8 0

3 years ago

Helen practiced the piano 5 days last week. The list below shows the number of minutes she practiced each day: 20,25,30,20,30 Wh

LekaFEV [45]

Answer:

4

Step-by-step explanation:

Given the data:

X : 20,25,30,20,30

Mean absolute deviation :

Σ(x - xbar) / n

n = sample size = 5

xbar = ΣX / n = (20+25+30+20+30) /5

xbar = 25

((20-25) + (25-25) + (30-25) + (20-25) + (30-25)) / 5

We take the absolute values, hence, negative signs are ignored :

(5 + 0 + 5 + 5 + 5) / 5

20 / 5

= 4

8 0

2 years ago

Slope= -2, y-intercept=5

vlabodo [156]

Answer:

y = -2x+5

Step-by-step explanation:

The formula for slope is y=mx+b.

(m = slope and b = y-intercept)

Plug in the numbers and you get your equation.

6 0

3 years ago

Can someone help me with this

irakobra [83]

Treat each side as if it's their own equation.
14-2*2
First multiply
14-4
Now subtract
10/5
Now divide
2
2 is the answer

5 0

3 years ago