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Romashka-Z-Leto [24]
3 years ago
8

Please help with an explanation​

Mathematics
1 answer:
Vsevolod [243]3 years ago
8 0

Answer:

the answer is incorrect though  the first step is correct but the second step is wrong and so is the answer. the correct answer would be x raise to the power of 2by 5

Step-by-step explanation:

{x⁶/⁵⁻²/⁵} ¹/²= {x⁴/⁵}¹/²= x²/⁵

in the second step the power of the denominator must be subtracted from the power of the numerator using the law of indices then we would get the correct answer

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If U = {a, b, c, d, e, f, g, h}, A= {a, b, c, d, e) and B= {c, d, e, f, g), find A-B and A nB. ​
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<em>here</em><em>,</em>

U = {a, b, c, d, e, f, g, h}, A= {a, b, c, d, e) and B= {c, d, e, f, g),

now,

A-B = {a,b}

A n B = { c,d,e}

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Read 2 more answers
National data indicates that​ 35% of households own a desktop computer. In a random sample of 570​ households, 40% owned a deskt
Elan Coil [88]

Answer:

Yes, this provide enough evidence to show a difference in the proportion of households that own a​ desktop.

Step-by-step explanation:

We are given that National data indicates that​ 35% of households own a desktop computer.

In a random sample of 570​ households, 40% owned a desktop computer.

<em><u>Let p = population proportion of households who own a desktop computer</u></em>

SO, Null Hypothesis, H_0 : p = 25%   {means that 35% of households own a desktop computer}

Alternate Hypothesis, H_A : p \neq 25%   {means that % of households who own a desktop computer is different from 35%}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p = sample proportion of 570​ households who owned a desktop computer = 40%

            n = sample of households = 570

So, <u><em>test statistics</em></u>  =  \frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }

                               =  2.437

<em>Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.</em>

Therefore, we conclude that % of households who own a desktop computer is different from 35%.

3 0
3 years ago
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