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liq [111]
3 years ago
15

Find the mean of the following data set 73, 81, 51, 76, 89, 95, 88

Mathematics
2 answers:
kifflom [539]3 years ago
6 0

Answer:

mean =  \frac{(73 + 81 + 51 + 76 + 89 + 95 + 88)}{7}  \\  = 79

Darya [45]3 years ago
6 0

Answer:

79

Step-by-step explanation:

What is mean?  The mean is the same as an average. To find the mean, add all the letters together. Then, divide this sum by the total amount of numbers.

Step One: 73+ 81+51 +76+ 89+ 95+88= 553

Step Two: Divide by numbers, in this case, there are 7: 553/7= 79.

There you go!

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Solve the following system of equations by the substitution method.
Dahasolnce [82]

Answer:

Step-by-step explanation:

5x=y+6   --------- equ 1

2x-3y=4  -------- equ 2

5x - 6 = y

y = 5x - 6

substitute in equ 2

2x-3(5x - 6) =4

2x - 3*5x + 3*6 = 4

2x - 15x + 18 = 4

- 13x = 4 - 18

-13x = -14

x = -14/-13

x = 14/13

substitute  in equ 1

5*14/13 = y+6  

70/13 - 6 = y

y =  ( 70 - 78)/13

y = -8/13

6 0
3 years ago
Karissa begins to solve the equation (x – 14) + 11 = x – (x – 4). Her work is correct and is shown below. (x – 14) + 11 = x – (x
sdas [7]
Her work is incorrect because she accidentally changed -14 to -7 in the second step.
Working it out from the top we get
(x-14)+11=x-(x-4)
x-14+11=x-x+4
x-3=0+4
Final answer:
x=7
Hope I helped :)
4 0
3 years ago
Read 2 more answers
I NEED HELP ON C,E,F,G PLEASE ASAP!!!!
Murrr4er [49]
Answer:
C. 364.4
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6 0
3 years ago
Let X1 and X2 be independent random variables with mean μand variance σ².
My name is Ann [436]

Answer:

a) E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

b) Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

Step-by-step explanation:

For this case we know that we have two random variables:

X_1 , X_2 both with mean \mu = \mu and variance \sigma^2

And we define the following estimators:

\hat \theta_1 = \frac{X_1 + X_2}{2}

\hat \theta_2 = \frac{X_1 + 3X_2}{4}

Part a

In order to see if both estimators are unbiased we need to proof if the expected value of the estimators are equal to the real value of the parameter:

E(\hat \theta_i) = \mu , i = 1,2

So let's find the expected values for each estimator:

E(\hat \theta_1) = E(\frac{X_1 +X_2}{2})

Using properties of expected value we have this:

E(\hat \theta_1) =\frac{1}{2} [E(X_1) +E(X_2)]= \frac{1}{2} [\mu + \mu] = \mu

So then we conclude that \hat \theta_1 is an unbiased estimator of \mu

For the second estimator we have:

E(\hat \theta_2) = E(\frac{X_1 + 3X_2}{4})

Using properties of expected value we have this:

E(\hat \theta_2) =\frac{1}{4} [E(X_1) +3E(X_2)]= \frac{1}{4} [\mu + 3\mu] = \mu

So then we conclude that \hat \theta_2 is an unbiased estimator of \mu

Part b

For the variance we need to remember this property: If a is a constant and X a random variable then:

Var(aX) = a^2 Var(X)

For the first estimator we have:

Var(\hat \theta_1) = Var(\frac{X_1 +X_2}{2})

Var(\hat \theta_1) =\frac{1}{4} Var(X_1 +X_2)=\frac{1}{4} [Var(X_1) + Var(X_2) + 2 Cov (X_1 , X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_1) =\frac{1}{4} [\sigma^2 + \sigma^2 ] =\frac{\sigma^2}{2}

For the second estimator we have:

Var(\hat \theta_2) = Var(\frac{X_1 +3X_2}{4})

Var(\hat \theta_2) =\frac{1}{16} Var(X_1 +3X_2)=\frac{1}{4} [Var(X_1) + Var(3X_2) + 2 Cov (X_1 , 3X_2)]

Since both random variables are independent we know that Cov(X_1, X_2 ) = 0 so then we have:

Var(\hat \theta_2) =\frac{1}{16} [\sigma^2 + 9\sigma^2 ] =\frac{5\sigma^2}{8}

7 0
3 years ago
What is the halfway point between 12 and 13?
nirvana33 [79]
12.5 is the halfway point
6 0
3 years ago
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