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2 years ago

8

When sample size increases Group of answer choices

1 answer:

Olenka [21]2 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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Find the height of this triangle.

Bogdan [553]

Answer:

$\sqrt{3}$

Step-by-step explanation:

x^2 + 1 = 4

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3 years ago

Pleaseeee helpppp !!!!!!!!! Will mark Brianliest !!!!!!!!!!!

Semenov [28]

Answer:

KL = 50

Step-by-step explanation:

∆JML is similar to ∆JNL. it follows that:

[tex] \frac{JM}{JN} = \frac{JL}{JK} [\tex]

JM = 4 + 20 = 24

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Plug in the values

[tex] \frac{24}{4} = \frac{10 + KL}{10} [\tex]

[tex] 6 = \frac{10 + KL}{10} [\tex]

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[tex] 6*10 = \frac{10 + KL}{10}*10 [\tex]

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2 years ago

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o-na [289]

The answer would be -40 because (-4)*(-2)=8. Then 8*(-5)=-40

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2 years ago

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The function M(s) = 225 + 0.65s represents the material cost of manufacturing gardening scissors when s scissors are produced. T

Llana [10]

Answer:

$\frac{279 + 1.8s}{s}$

Step-by-step explanation:

The material cost of manufacturing gardening scissors when s scissors are produced : $M(s) = 225 + 0.65s$

The labor cost for producing s scissors: $L(s) = 54 + 1.15s$

So, total cost of producing s scissors = $M(s)+L(s)$

= $225 + 0.65s+54 + 1.15s$

= $279 + 1.8s$

So, total cost for s scissors = $279 + 1.8s$

So, manufacturing cost for 1 scissor= $\frac{279 + 1.8s}{s}$

Hence the manufacturing cost per scissors is $\frac{279 + 1.8s}{s}$

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2 years ago

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Setler [38]

I hope this helps you

x^2+2.(-3).x+(-3)^2

(x-3).(x-3)

(x-3)^2

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3 years ago