Ask question

Ask question

All categories

3 years ago

12

Tony sees 28 patients in 7 days at his clinic. Jeff sees 100 patients in 25 days at his clinic. Assuming that each doctor sees p

atients at a constant rate, which conclusion can you make about the rate of service for these doctors?

1 answer:

sammy [17]3 years ago

3 0

28/7 = 4 patients a day.

100/25 = 4 patients a day.

Therefore, you can conclude that doctors see 4 patients each day at the clinic.

You might be interested in

Mr. brown sells horse supplies. A pair of riding gloves sells for $16. He says he will make $144 if he sells 9 pairs is it reaso

kap26 [50]

16 X 9 = 144. Yes this is completely reasonable.

6 0

3 years ago

Read 2 more answers

A zookeeper weighed an African elephant to be 9 × 103 pounds and an African lion to be 4 × 102 pounds. How many times greater is

amm1812

Answer:

The African Elephant is 2.27 times heavier than a African Lion.

Step-by-step explanation:

9 × 103 pounds = 927E.

4 × 102 pounds = 408L.

4 0

3 years ago

Read 2 more answers

PLEASE HELP .!!! ILL GIVE BRAINLIEST.. *EXRTA POINTS* .. DONT SKIP :(( ! <br> ILL GIVE 30 POINTS .

Rufina [12.5K]

Answer:

D

Step-by-step explanation:

they don't eat the same food, so it rules out the first three sorry if wrong

4 0

3 years ago

Read 2 more answers

Karen is saving money to buy a game. The game costs $12 , and so far she has saved two-thirds of this cost. How much money has K

Mashutka [201]

$\dfrac{2}{3}\cdot\$12=2\cdot\$4=\$8$

3 0

3 years ago

Read 2 more answers

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