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3 years ago

Please help! Thanks!: If a 95% confidence interval was being calculated using the following expression:

Mathematics

1 answer:

Roman55 [17]3 years ago

4 0

The confidence interval can be calculated from the margin of error, and vice versa,

The margin of error of the confidence interval is 2.156

The confidence interval is given as:

$\mathbf{CI = 3.2 \pm 1.96 \cdot 1.1}$

The formula of confidence interval is:

$\mathbf{CI = \bar x\pm E}$

Where E represents the margin of error.

So, by comparison:

$\mathbf{E = 1.96 \cdot 1.1}$

Multiply

$\mathbf{E = 2.156}$

Hence, the margin of error of the confidence interval is 2.156

Read more about margin of errors at:

brainly.com/question/13990500

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Someone help please and thank you!!

svlad2 [7]

Https://www.symbolab.com/solver/function-asymptotes-calculator this is a good asymptote calculator

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3 years ago

Using Euclid's division algorithm prove that : 847,2160 are co-primes

AysviL [449]

Definition of eo-primes or relatively primes: Two numbers are said to be co-prime or relatively prime If their HCF IS 1 Hence to prove 847 and 2160 as co-prime numbers we will find their HCF and which should be 1

New steps to find HCF will be as under

2160 = 847 x 2+ 466
847 = 466 ×1 +381
466 = 381 x 1 + 85
381 =85 x 4+ 41
85 =41 x 2+3
41 =3 x 13+ 2
3 =2 x 1+1
2 =1 x 2+0

Therefore, the HCF=1 Hence, the numbers are co-primes (relatively prime).

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3 years ago

I really need help pleaseeee

raketka [301]

12.92 I hope this helps

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2 years ago

Bonjour,est ce que vous pouvez me faire cette exo svp

saveliy_v [14]

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3 years ago

Javier simplified the expression below. Find and describe the three mistakes he made.

OverLord2011 [107]

$( \dfrac{20x}{5x^8})^-^3$

<span>Take out the constants
</span>
$( \dfrac{20}{5} * \dfrac{x}{x^8} ) ^-^3$

$(4*\dfrac{x}{x^8} ) ^-^3$

Use the rule : $\frac{{x}^{a}}{{x}^{b}}={x}^{a-b}
$

$(4x^1^-^8)^-^3$

$(4x^-^7)^-^3$

Use the n<span>egative power rule : </span> ${x}^{-a}=\frac{1}{{x}^{a}}
$

$\frac{1}{4(x^-^7)}^-^3$

$\frac{1}{4^3(x^-^7)}^-^3$

$\dfrac{1}{64x^-^2^1}$

$\dfrac{1}{64* \dfrac{1}{x^2^1}}$

$\dfrac{1}{64 \dfrac{}{x^2^1}}$

$= \dfrac{x^2^1}{64}$

6 0

3 years ago

Read 2 more answers