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3 years ago

When sample size increases Group of answer choices

Mathematics

1 answer:

Olenka [21]3 years ago

3 0

Answer:

Option A)

Confidence interval decreases

Step-by-step explanation:

If we increase the sample size, then,

The standard error of the interval decrease.
If the standard error increase, the margin of error of the interval decrease.
If the margin of error decreases, the width of the confidence level decreases, hence, the confidence interval become narrower.

Thus, the correct answer is

Option A)

Confidence interval decreases

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at a checkers tournament there where 64 players. It was elimatation tournament. how many games where played before there was a w

Inessa05 [86]

So first game
half were eliminated
we notice a pattern
first game=64 times 1/2
next game 32 times 1/2 so

so just multiply by 1/2 till you get 1
round 1 64/2=32
round 2 32/2=16
round 3 16/2=8
round 4 8/2=4
round 5 4/2=2
round 6 2/2=1

6 is the answer

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4 years ago

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Simplify the expression show all of your work 3+5 squared -9÷3×4= Please help

dexar [7]

Answer:

Step-by-step explanation:

(5√3*√3)+(5√3*5)+(-1*√3)+(-1*5)

5*3+25√3-√3-5

15+24√3-5

**24√3+10**

Then to solve the second, apply division rules within the radical. This means you can cancel an m^1 and n^4 from the bottom and top of the fraction. This leaves...

3√(88m^19*n^8)

(That might be all you need to do, otherwise you can take the square root of each number in the term giving...

3(√88)*m^9.5*n^4)

Hope I made it clear enough

Please give me Brainliest

7 0

4 years ago

Read 2 more answers

The following describes a sample. The information given includes the five number summary, the sample size, and the largest and s

lys-0071 [83]

Answer:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

Step-by-step explanation:

For this case we have the 5 number summary

(160,210,220,250,270)

So then we have:

minimum = 160 , Q1 = 210, Q2= Median=220, Q3 = 250, Max=270

If we find the interquartile range we got:

$IQR = Q_3 -Q_1 = 250-210 =40$

For this case we need to find the lower and upper limit with the following formulas:

$L= Q_1 - 1.5*IQR = 210 -1.5*40 =150$

$U= Q_3 + 1.5*IQR = 250 +1.5*40 =310$

And if we analyze the info provided we have 500 values

Tails: 160,165,167,171,175,...,268,269,269,269,270,270

So as we can see on the tails any values is <150 or >310 so then for this case we cannot consider as outliers any of the values on the tails of the distribution.

4 0

4 years ago

Help pls with this problem

Aleonysh [2.5K]

Answer:

x=45

Step-by-step explanation:

180-32=148

148/2=74

180-74=105

105-60=45

x=45

3 0

3 years ago

A ball bounces back 0.6 of its height on every bounce. If a ball is dropped 200 feet, how high does it bounce the fifth bounce?

MissTica

It is asking you to slice off 40% of the bounce height in each subsequent bounce.

So you start with full height 1, that's n = 0 bounces. Then after n = 1 bounce the new height is .6. After the second bounce, n = 2, the height becomes .6 X .6 = .36. And after n = 3, we have .6 X .6 X .6 = .216

So by induction we see that h(n = 5) = .6^n = .6^5 = 0.07776 = .078 of its original height or H = .078*200 = 15.6 ft. after five bounces. ANS. 16.0 if rounded to the nearest tenth

4 0

4 years ago