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3 years ago

Tell me about what you know about "Rounding"

Mathematics

2 answers:

miv72 [106K]3 years ago

8 0

Answer:

Rounding is about estimating

Step-by-step explanation:

Rounding is finding the number that it is close to. It makes solving problems easier and once you understand it, it helps A LOT with estimating.

user100 [1]3 years ago

4 0

It is finding the closest estimate to number. Rounding is used in everyday calculations and in many different fields of work

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Brian needs to paint a logo using two right triangles. The dimensions of the logo are shown below. What is the difference betwee

solong [7]

Answer:

7.5 cm²

Step-by-step explanation:

Dimensions of the large ∆:

$base (b) = 3cm, height (h) = 9cm$

$Area = 0.5*b*h = 0.5*3*9 = 13.5 cm^2$

Dimensions of the small ∆:

$base (b) = 2cm, height (h) = 6cm$

$Area = 0.5*b*h = 0.5*2*6 = 6 cm^2$

Difference between the area of the large and the small ∆ = 13.5 - 6 = 7.5 cm²

5 0

3 years ago

find two consecutive odd numbers such that the sum of the larger number and twice the smaller number is 27 less then four times

soldier1979 [14.2K]

N is an odd integer

The Next Larger is N+2

N+2 + 2N = -27 + 4N

3N + 2 = 4N - 27

0 = 4N - 27 - 3N - 2

0 = N - 29

N = 29

They are 29 and 31.

31 + 2(29) = 4(29) - 27

31 + 58 = 89 = 116 - 27 yes it works

4 0

3 years ago

What is the point-slope form for the line passing through (3, -6) and (-1, 2)

Leto [7]

Answer:

y-2 = -2(x+1)

Step-by-step explanation:

Find the slope(m) first using the two points (3, -6) and (-1, 2).

Slope: -2

Next, plug in x₁ and y₁. I plugged in (-1, 2)

5 0

3 years ago

PLEASE HELP

LekaFEV [45]

Answer: one solution

Step-by-step explanation:

CONCEPT:

- One solution is when the final variable would be able to find a solution.

- Infinite solution is when the equations result in 0=0, meaning any number will fit

- No solution is when the equation results in both sides on equal.

SOLVE:

Given

3/4x-7=3(1/6x+1)

Expand Parenthesis

3/4x-7=1/2+3

Multiply both sides by the LCM of fractions (Least Common Multiple)

4(3/4x-7)=4(1/2x+3)

3x-28=2x+12

Subtract both sides by 2x

3x-28-2x=2x+12-2x

x-28=12

Add both sides by 28

x-28+28=12+28

x=40

Since we are able to get exactly one solution, there is only one solution.

Hope this helps!! :)

Please let me know if you have any quesionts

5 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