All categories

3 years ago

12. What is the effect when the sample size is

Mathematics

1 answer:

katrin [286]3 years ago

3 0

Confidence interval decreases, error decreases

Option: a

Explanation:

Increasing of sample size will decrease the width of the confidence intervals because it also decreases the standard error of the interval. So whenever the 'sample size' is increases then the 'confidence interval' decreases as well as the 'error' decreases.

Confidence interval gives us the unknown average population or percentage of the population. This gives us just how much the average rate is varying. Confidence interval gives the location as well as the precision of a measure. The smaller sample sizes gives more wider intervals.

You might be interested in

From a fraction to a decimal 19/20

REY [17]

It would be 0.95 you judt divide the top jumberto thebottom one

5 0

3 years ago

PLEASE ANSWER WITH AN EXPLANATION! THANK YOU

lubasha [3.4K]

Answer:

$\large\boxed{A=153\ cm^2}$

Step-by-step explanation:

Look at the picture.

We have

square with side length a = 9

trapezoid with base lengths b₁ = 9 and b₂ = 6 and the height length h = 6

right triangle with legs lengths l₁ = 3 + 6 = 9 and l₂ = 6

The formula of an area of a square

$A=a^2$

Substitute:

$A_I=9^2=81\ cm^2$

The formula of an area of a trapezoid:

$A=\dfrac{b_1+b_2}{2}\cdot h$

Substitute:

$A_{II}=\dfrac{9+6}{2}\cdot6=\dfrac{15}{2\!\!\!\!\diagup_1}\cdot6\!\!\!\!\diagup^3=(15)(3)=45\ cm^2$

The formula of an area of a right triangle:

$A=\dfrac{l_1l_2}{2}$

Substitute:

$A_{III}=\dfrac{(9)(6)}{2}=\dfrac{54}{2}=27\ cm^2$

The area of the shape:

$A=A_I+A_{II}+A_{III}\\\\A=81+45+27=153\ cm^2$

8 0

3 years ago

Jean and Jericho who are playing in the school grounds decided to sit on a

Natalija [7]

We want to see underline the correct part in each statement.

1) This situation illustrates (direct, inverse) variation.
2) The two quantities that must vary in this situation are (weight and 
distance from the center, height and weight).
3) The heavier the kid, the (closer, farther) he/she should be at the
center to balance the seesaw?
4) When Jean moves farther from the center, Jericho tends to go (up,
down).
5) If Jericho moves closer to the center, Jean tends to go (up, down).

So, Jean and Jericho are playing on a seesaw.

Jean is heavier than Jericho.

Now, notice that a seesaw is a lever. So it "amplificates" the force that you apply in one end to lift the weight that is on the other end. Depending on the form of the lever and the weights, the force that you need to do changes.

If we define:

W₁ = Jean's weight.
d₁ = Jean's distance to the center.
W₂ = Jericho's weight.
d₂ = Jericho's distance to the center.

We must have:

W₁*d₁ = W₂*d₂

Then:

1) This situation illustrates (direct, inverse) variation.

d₁, the position of Jean, varies inversely with respect to Jean's weight.

2) The two quantities that must vary in this situation are (weight and

distance from the center, height and weight).

(by the equation above)

3) The heavier the kid, the (closer, farther) he/she should be at the

center to balance the seesaw?

By the given equation, we see that d₁ must be smaller than d₂.

And the last two are trivial:

4) When Jean moves farther from the center, Jericho tends to go (up,

down).

5) If Jericho moves closer to the center, Jean tends to go (up, down).

If you want to learn more, you can read:

brainly.com/question/18320907

7 0

2 years ago

Find the area of a polygon with the given vertices A(2,2) B(5,2) C(3,-1) D(0,-1) Number 21

Bogdan [553]

Answer:

Je pense que je ne sais pas vraiment

6 0

3 years ago

Which cost less a 1,000 dollar computer discounted onto 20% then 10% or a 1,000 dollar computer discounted to 30% if that makes

kykrilka [37]

First one costs $720, second costs $700 so the second one the 1,000 dollar computer discounted to 30%

6 0

3 years ago