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Reptile [31]
3 years ago
15

What is the approximate area of a circle with a diameter of 28 centimeters? Use π = 3.14.

Mathematics
2 answers:
larisa [96]3 years ago
8 0
Formula is π * r^2

π = about 3.14 and the radius is half of the diameter so 28/2 = 14

3.14 * 14^2 = 615.44

B

Nikitich [7]3 years ago
3 0

Answer:

B.

615.4 square centimeters

Step-by-step explanation:

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The dot plot represents the prices of different brands of chocolate bars.
Gala2k [10]
The answer is simply C.$2.50
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3 years ago
Read 2 more answers
1. Which would be a more helpful for solving the system: adding the two equations or subtracting
ira [324]

Answer:

1) Subtracting one from the other

2) x = 0.375 and y = 0.125

Step-by-step explanation:

Equation 1 :

\frac{2x+1}{2y}=7\\\\

Equation 2:

\frac{6x-1}{2y}=5

To solve these equations by the Elimination method we multiply equation 1 with 6 and multiply equation 2 with 2 so now,

\frac{2x+1}{2y} =7\\\\2x+1=14y\\\\Multiplying\ Equation\ 1\  with\ 6\\\\12x+6=84y

Now for the second equation:

\frac{6x-1}{2y}=5\\\\6x-1=10y\\\\Multiplying\ equation\ 2\ with\ 2 \\\\12x-2=20y

Now subtracting equation 2 from equation 1

12x+6=84y\\\\12x-2=20y\\\\Subtracting\ leads\ to\\12x-12x+6-(-2)=84y-20y\\0+8=64y\\8=64y\\8/64=y\\0.125=y

now insert this value of y into any equation

lets insert it into equation 1

\frac{2x+1}{2y} =7\\\\2x+1=14y\\2x+1=14(0.125)\\2x+1=1.75\\2x=1.75-1\\2x=0.75\\x=0.75/2\\x-0.375

5 0
3 years ago
Which triangle side lengths form a right triangle? Choose all that apply.
Radda [10]

Your answer would be C (8,15,17) and E (15,20,15) or is it 25?

Step-by-step explanation:

:)

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
3 years ago
Complete the equation for f(x)
I am Lyosha [343]

Answer:

f(x)=9

Step-by-step explanation:

8 0
3 years ago
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