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3 years ago

-4+2(-2+5)^2 ______________ 5-(2-4)^3

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

8 0

Answer:

32/13

Step-by-step explanation:

Please include the instructions when you post a problem.

Let's assume you re to simplify this expression.

We get:

-4+2(-2+5)^2 -4 + 2(3)^2 -4 + 36

__________ = ------------------- = ----------------- = 32/13

5-(2-4)^3 5 - (-2)^3 5 + 8

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5x +6y = 20 (1)

slega [8]

(2) + (2)

=> 8x - 6y = -46 ———(3)

(1) + (3)

=> 5x + 6y + (8x - 6y) = 20 - 46

=> 13x + (6-6)y = -26

=> 13x + 0y = -26

=> 13x = -26

=> x = -26/13

=> x = -2

=> 5(-2) + 6y = 20

=> 6y = 20 + 10

=> 6y = 30

=> y = 5

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A sample size must be determined for estimating a population mean given that the confidence level is % and the desired margin o

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Complete Question

The complete question is shown on the first uploaded image

Answer:

$n =290$

$n =129$

The correct option is D

Step-by-step explanation:

From the question we are told that

The margin of error is $E = 0.23$

The largest value is k = 15

The smallest value is u = 7

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally population standard deviation estimate is mathematically represented as

$\sigma = \frac{Range }{4}$

Here the Range is mathematically represented as

$Range = k - u$

=> $Range = 15 - 7$

=> $Range = 8$

=> $\sigma = \frac{8 }{4}$

=> $\sigma = 2$

Generally the sample size is mathematically represented as

$n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2$

=> $n = [1.96 * 2 }{0.23} ] ^2$

=> $n =290$

Generally to obtain a conservatively small sample size the population standard deviation estimate becomes

=> $\sigma = \frac{Range }{6}$

=> $\sigma = \frac{8 }{6}$

=> $\sigma =1.333$

Generally the conservatively small sample size is mathematically evaluated as

$n = [\frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ] ^2$

=> $n = [1.96 * 1.333 }{0.23} ] ^2$

=> $n =129$

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