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3 years ago

What is the product of 3 × 50 × 18 × 2? A. 2,700 B. 540 C. 1,800 D. 5,400

Mathematics

2 answers:

ozzi3 years ago

4 0

Answer:

The answer is D

Step-by-step explanation:

Hope I helped!! :)

sp2606 [1]3 years ago

3 0

The answer is D because when you insert it into a calculator it gives you D. Yet, if you do it manually you still get the same number .

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Write me What is 10/12 using simplest form

Serjik [45]

Found it online hope that’s helps

8 0

3 years ago

Read 2 more answers

Will give brainlist Very unlikely answer choices: A. 1 B. 0.01 C. 0.9 D. 0 E. 0.6 F. 0.5 G. 0.35

emmasim [6.3K]

Answer:

B) 0.01

Step-by-step explanation:

If they want to know which one is very unlikely, you could turn them into percents.

1=100%

0.01=1%

0.9=90%

0=0%

0.6=60%

0.5=50%

0.35=35%

Since 0% is "impossible", not just "very unlikely", I'd go with 1%, which is B) 0.01

This is assuming I'm understanding your question correctly, since it doesn't actually ask a question.

5 0

3 years ago

my brother and i walk the same route every day. my beother takes 40minutes to get to school and i take 30minutes to get to schoo

Goryan [66]

In 1 minute, I walk = 1/30

In 1 minute, My brother walk = 1/40

In T minutes, I walk = T*1/30

My brother walk 8 minute before = T + 8

In 1 minute, My brother walk = (T + 8)*1/40

When these two distances are same, then I will catch him.

T*(1/30) = (T + 8)*(1/40)

30*(T + 8) = 40*T

30*T + 240 = 40*T

240 = 10*T

T = 24 minutes

hence, I will catch him in 24 minutes.

Take 5 instead of 8.

My brother walk 5 minute before = T + 5

In 1 minute, My brother walk = (T + 5)*1/40

When these two distances are same, then I will catch him.

T*(1/30) = (T + 5)*(1/40)

30*(T + 5) = 40*T

30*T + 150 = 40*T

150 = 10*T

T = 15 minutes

Hence, I will catch him in 15 minutes.

6 0

3 years ago

5 miles to 35 miles in a fraction ?

LenKa [72]

5/35
Simplify by dividing both sides by 5
1/7
Final Answer: 1/7

4 0

2 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