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konstantin123 [22]

3 years ago

11

After paying eight dollars for the pie, Mike has eighty - one dollars left, his friend has thirteen dollars. How much money did

he have before buying the pie

1 answer:

Inessa [10]3 years ago

7 0

Answer:

__ - 8=81

unsure if we're adding his friend but for this i assume no,

81+ 8 = 89

so its either 89 or

89+13=102

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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

Jessica plans to purchase a car in one year at a cost of $30,000. how much should be invested in an account paying 10% compounde

dezoksy [38]

The formula is
A=p (1+r/k)^kt
A fund needed 30000
p Amount invested?
R interest rate 0.1
K compounded semiannual 2
T time 1 year
Solve the formula for p to get
P=A÷(1+r/k)^kt
P=30,000÷(1+0.1÷2)^(2×1)
P=27,210.88

4 0

3 years ago

The local Tupperware dealers earned the following commissions last month. What was the mean commission earned? Round your answe

kirza4 [7]

Answer:

B. $ 3094.01

Step-by-step explanation:

Given observations are,

$2894.21, $1777.15, $2144.77, $4096.37, $4046.29, $1786.37, $3296.69, $4086.27, $2784.22

Number of observations = 10,

Sum of the observations = 2894.21+1777.15+2144.77+4096.37+4046.29+1786.37+3296.69+4086.27+2784.22+4027.79 = 30940.13,

Hence,

$Mean = \frac{\text{Sum of observations}}{\text{Number of observation}}$

$=\frac{30940.13}{10}$

$=3094.013$

$\approx 3094.01$

Option 'B' is correct.

5 0

3 years ago

Sarah has a bag of 50 skittles, some are red and some are yellow. Sarah randomly pulls a skittle out of the bag and records it,

Leno4ka [110]

Answer:

20

Step-by-step explanation:

Assume n is the number of times she pick a skittle out of the bag

The probability she pick out red : 8/n

The probability she pick out yellow : 12/n

=> The ratio of red and yellow skittle : 8/n : 12/n = 8/12 = 2/3, means every 2 red skittles we have 3 yellow ones.

But we have 50 skittles, so the likely numbers of reds in the bag is = 2* $\frac{50}{5}$ = 20

6 0

3 years ago

23 is what percent of 115<br> (Explain Steps)

Sholpan [36]

Answer:

20%

Step-by-step explanation:

23/115=.2

8 0

2 years ago