2 years ago

Find the quotient of -88,11

Mathematics

1 answer:

nata0808 [166]2 years ago

7 0

The answer is 11....................

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Let A,B, and C be square invertible matrices of the same size. If C has no eigenvalue equal to -1, then (AB + ACB)^-1 is equal t

Kisachek [45]

AB + ACB = A (B + CB) = A (I + C ) B

Taking the inverse gives

(A (I + C ) B)⁻¹ = B ⁻¹ (I + C )⁻¹ A ⁻¹

so the answer is (A)

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2 years ago

What is the fractional equivalent of the repeating decimal n=0.2727…?

erastova [34]

Answer: 3/11 would be the answer

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tresset_1 [31]

F( x ) = 0 => / x / = 0 => x = 0, where -2 <= 0 < 3 ;
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2 years ago

Which is the best estimate for 19% of 205? • A. 40 • B. 50 • C. 100 • D. 400

adelina 88 [10]

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Step-by-step explanation:

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2 years ago

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In a sample of 41 temperature readings taken from the freezer of a restaurant, the mean is 29.7 degrees and the population stand

yKpoI14uk [10]

Answer: $=(29.1596\text{ degrees},\ 30.2404\text{ degrees})$

Step-by-step explanation:

Given: Sample size: n = 41

Sample mean $\overline{x}= 29.7$ degrees

Population standard deviation $\sigma=2.7$ degrees

Confidence level (c) = 80% =0.80

Significance level (a) = 1- c = 1-0.80 = 0.20

z-score for 80% confidence level : z = 1.2816 [from z-table]

Confidence level for population mean :-

$\overline{x}\pm z\dfrac{\sigma}{\sqrt{n}}$

$=29.7\pm ( 1.2816)\dfrac{2.7}{\sqrt{41}}$

$=29.7\pm ( 1.2816)\dfrac{2.7}{6.403124}$

$=29.7\pm ( 1.2816)(0.42167)$

$=29.7\pm 0.5404$

$=(29.7-0.5404,\ 29.7+0.5404)$

$=(29.1596,\ 30.2404)$

Hence, 80% confidence interval for the temperatures in the freezer $=(29.1596\text{ degrees},\ 30.2404\text{ degrees})$

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2 years ago