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2 years ago

Which of the following is an equation of a line that is parallel to one whose equation is y = 5x - 3?

Mathematics

1 answer:

LekaFEV [45]2 years ago

3 0

Any line that has 5 as its slope. So which ever one says “5x”

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Help me please I’m so confused

lesya [120]

12= -12
-1/2= 1/2
.25 =-.25
-27=27
0=0

8 0

3 years ago

Giving brainliest for correct awnser. Factor the expression below.

tresset_1 [31]

Answer:

B. (x + 3) (x + 27)

Step-by-step explanation:

To factor the expression, you must split the middle term (18x) into two terms that can be added to get 18, and multiplied to get 81:

$x^2 + 18x + 81$

$x^2 +3x + 27x + 81$

Group:

$(x^2 +3x) (27x + 81)$

Take out GCF (Greates Common Factor):

x(x + 3) 27(x + 3)

(x + 27) (x + 3)

8 0

2 years ago

Is it essary to rename 4 1/4 if you subtract 3/4 from it.

Archy [21]

No . because you would then have 2/4 which is 1/2 50%

3 0

3 years ago

15 is 30% of what number?

777dan777 [17]

15 is 30% of 50

Change the percentage to a decimal by dividing it over 100:
$\frac{30}{100} = 0.3$

Divide 15 over 0.3:
$\frac{15}{0.3} = 50$

3 0

3 years ago

Read 2 more answers

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