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3 years ago

Emily drives her car 42 miles in 1 hour. what is her average speed in miles per hour.

Mathematics

2 answers:

Svetradugi [14.3K]3 years ago

7 0

Answer:

65 mph would be the answer

Step-by-step explanation:

masha68 [24]3 years ago

4 0

Her average mpg would be 42mph. Seeing as she travels. 42 miles an hour

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Suppose you are to select a password of size 4. The first character should be any lowercase letter or a digit, while the second

Goshia [24]

First question seems incomplete :

Answer:

40 ways

Step-by-step explanation:

Question B:

Number of boys = 6

Number of girls = 4

Number of people in committee = 3

Number of ways of selecting committee with atleast 2 girls :

We either have :

(2 girls 1 boy) or (3girls 0 boy)

(4C2 * 6C1) + (4C3 * 6C0)

nCr = n! ÷ (n-r)!r!

4C2 = 4! ÷ 2!2! = 6

6C1 = 6! ÷ 5!1! = 6

4C3 = 4! ÷ 1!3! = 4

6C0 = 6! ÷ 6!0! = 1

(6 * 6) + (4 * 1)

36 + 4

= 40 ways

3 0

2 years ago

The table shows the height, in centimeters, that a weight bouncing from a spring would achieve if there were no friction, for a

Paha777 [63]

Hello! For ease of calculations, we can identify the time it took for the weight to bounce back to the other direction, then the other, and then back to its original position by looking at the time it took for the weight to change from 0 to 25 to 0 to -25 then back to 0. This is one whole cycle of the weight.

By the time the weight first reached zero, 1.5 seconds has passed. By the third time it got to zero again, 7.5 seconds has passed. Therefore, one whole cycle of the weight is 7.5-1.5 = 6.0 seconds.

ANSWER: One whole cycle of the weight took 6 seconds.

4 0

3 years ago

What is the value of x

Evgen [1.6K]

Answer:

Step-by-step explanation:

Given

with reference angle 45°

hypotenuse (h) = x

perpendicular (p) = 6√2

Now

with reference angle 45°

sin 45° = p / h

1 / √2 = 6√2 / x

Cross multiply ✖

6√2 * √2 = x

x = 6 * 2

x = 12

6 0

3 years ago

Mr.Brookins asked his students to plot an interger on a number line. Which two students could have represented the same number?

VikaD [51]

Answer:

I need a picture

Step-by-step explanation:

8 0

3 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