All categories

3 years ago

50 POINTS FOR THIS ALGEBRA WORD PROBLEM

Mathematics

2 answers:

olga55 [171]3 years ago

7 0

Let x,x+2,x+4,x+6 be 4 consecutive no then,

sum of first 3 no=4x+12

According to condition,

20 less than 5 times 4th= 4 more than sum of 1st,2nd and 3rd

5(x+6)-20=4+(sum of first 3 no)

5(x+6)-20=4+(4x+12)

5(x+6)-20=4x+16

5x+30-20=4x+16

5x+10=4x+16

5x-4x=16-10

x=6

So,

1st no=x=6

2nd no=x+2=8

3rd no=x+4=6+4=10

4th no=x+6=6+6=12

Dahasolnce [82]3 years ago

7 0

Answer:

Let x,x+2,x+4,x+6 be 4 consecutive no then,

sum of first 3 no=4x+12

According to condition,

20 less than 5 times 4th= 4 more than sum of 1st,2nd and 3rd

5(x+6)-20=4+(sum of first 3 no)

5(x+6)-20=4+(4x+12)

5(x+6)-20=4x+16

5x+30-20=4x+16

5x+10=4x+16

5x-4x=16-10

x=6

So,

1st no=x=6

2nd no=x+2=8

3rd no=x+4=6+4=10

4th no=x+6=6+6=12

Step-by-step explanation:

You might be interested in

Ba + 3ca = -2 Solve for a

monitta

We have:

ba + 3ca = -2

Extract common factor "a":

a(b + 3c) = -2

Divide by the term (b + 3c):

$a = \dfrac{-2}{b+3c}$

3 0

3 years ago

Describe the differences between vertical and adjacent angles

Sunny_sXe [5.5K]

Vertical angles are pairs of opposite made by 2 intersecting lines
and adjacent lines are 2 angles that have a common endpoint and common sides

8 0

3 years ago

Please help 10 point reward

LekaFEV [45]

ANSWER: I can give you the steps to figure out this question.

1 if it is starting at the 6 on the y axis, then just do up 1 over 1 until you get to the 10

The answer to your question is..... 6y+1=4x+10

4 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

3 years ago

A crane lifts a 425 kg steel beam vertically a distance of 66 m. How much work does the crane do on the beam if the beam acceler

Fiesta28 [93]

The solution to the problem is as follows:

W=Fd=mad=425*1.8*66 =50490 J

Therefore, it takes 50490 J of work the crane will do to accelerate the beam upward at 1.8 m/s^2.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

6 0

3 years ago

Read 2 more answers