All categories

3 years ago

A car travels at an average speed of 52 miles per hour How many miles does it travel in 5 hours and 45 minutes

Mathematics

2 answers:

Gre4nikov [31]3 years ago

6 0

First of all, lets convert time 5 hrs 45 minutes in single unit. (in hrs only)

time : 5 hrs + 45/60 hrs = 5 + 0.75 = 5.75 hrs

distance = time * speed

distance = 5.75 * 52 = 299 miles

hammer [34]3 years ago

3 0

5 h 45 min = 5.75 hours

52 * 5.75 = 299 miles

You might be interested in

A salesman's commission rate is 6%. What is the commission from the sale of 36,000 worth of furnaces?

borishaifa [10]

Answer:

2160

Step-by-step explanation:

36000*6/100=2160

6 0

3 years ago

Arnold's entire workout consisted of 10 minutes of warm up exercise, 25 minutes of lifting weights, and 15 minutes on the treadm

lord [1]

Add all the times together:

10 + 25 + 15 = 50 minutes total.

For the ratio divide the time for weights by total time:

25/50 which reduces to 1/2

The ratio is 1/2

5 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

Help me quickly pls<3

Kipish [7]

Answer:

Step-by-step explanation:

No others make sense because you can't use multiplication in this problem and well it's hard to explain honestly lol. But I'm almost certain your answer is A

5 0

2 years ago

What is 12 * 12<br><br> i know the answer just to work your brains and give some points

AfilCa [17]

$12 \times 12 = 144$

Hope I work my brain.

3 0

2 years ago

Read 2 more answers