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Blababa [14]
3 years ago
15

Is it true that Confidence intervals are always close to their true population values?

Mathematics
1 answer:
user100 [1]3 years ago
6 0

Answer:

This means that there is a 95% probability that the confidence interval will contain the true population mean. Thus, P( [sample mean] - margin of error < μ < [sample mean] + margin of error) = 0.95.

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Which of the following represents the top view of the figure? PLZ HURRY
lana [24]
The third one. The one that goes long box, short, short, long. That's because the first box isn't covered, and they next two are about half boxes showing and the last one is a full box again. 
6 0
3 years ago
x, y, a, and b are positive integers. When x is divided by y, the remainder is 6. When a is divided by b, the remainder is 9. Wh
Mademuasel [1]

Answer:

y + b > 15

Step-by-step explanation:

given data

positive integers = x, y, a, b

x divided by y then remainder = 6

a divided by b then remainder = 9

to find out

possible value for y + b

solution

we know that here when  x divided by y then remainder is 6

its mean y is greater than 6 and when a divided by b then remainder is 9

so its mean b is greater than 9

and we know remainder is less than divisor

so here y + b must be greater than = 6 + 9

y + b > 15

5 0
3 years ago
What write a fraction that is a multiple of 4/5
ahrayia [7]
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5 0
3 years ago
Read 2 more answers
Can someone solve this
Natalka [10]

Option D:

ΔCAN ≅ ΔWNA by SAS congruence rule.

Solution:

Given data:

m∠CNA = m∠WAN and CN = WA

To prove that ΔCAN ≅ ΔWNA:

In ΔCAN and ΔWNA,

CN = WA (given side)

∠CNA = ∠WAN (given angle)

NA = NA (reflexive side)

Therefore, ΔCAN ≅ ΔWNA by SAS congruence rule.

Hence option D is the correct answer.

4 0
3 years ago
Which four inequalities can be used to find the solution to this absolute value inequality?
natita [175]

The four inequalities that can be used to find the solution of 3 ≤ |x + 2| ≤ 6 is x + 2 ≤ 6, x + 2 ≥ -6, x + 2 ≥ 3 and x + 2 ≤ -3

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more variables and numbers.

Given the inequality:

3 ≤ |x + 2| ≤ 6

Hence:

x + 2 ≤ 6, -(x + 2) ≤ 6, 3 ≤ x + 2 and 3 ≤ -(x + 2)

This gives:

x + 2 ≤ 6, x + 2 ≥ -6, x + 2 ≥ 3 and x + 2 ≤ -3

The four inequalities that can be used to find the solution of 3 ≤ |x + 2| ≤ 6 is x + 2 ≤ 6, x + 2 ≥ -6, x + 2 ≥ 3 and x + 2 ≤ -3

Find out more on equation at: brainly.com/question/2972832

#SPJ1

6 0
2 years ago
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