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cestrela7 [59]
3 years ago
13

Someone please help me

Mathematics
1 answer:
Anika [276]3 years ago
6 0

Answer:

A

Step-by-step explanation:

Given

a = \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy} ← as a single fraction

Thus

\frac{1}{a} = \frac{1}{\frac{y-x}{xy} } = \frac{xy}{y-x} → A

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Maslowich

Use a calculator and type

4.108 +

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19.338



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Brahmagupta solved a quadratic equation of the form ax2 + bx = c using the formula x =, which involved only one solution. Using
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Not sure about the first one, The second one is 1/9.

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3 years ago
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3 years ago
In 1943​, an organization surveyed 1100 adults and​ asked, "Are you a total abstainer​ from, or do you on occasion​ consume, alc
Fynjy0 [20]

Answer:

We conclude that the proportion of adults who totally abstain from alcohol​ has changed.

Step-by-step explanation:

We are given that in 1943​, an organization surveyed 1100 adults and​ asked, "Are you a total abstainer​ from, or do you on occasion​ consume, alcoholic​ beverages?"

Of the 1100 adults​ surveyed, 429 indicated that they were total abstainers. In a recent​ survey, the same question was asked of 1100 adults and 352 in

<u><em>Let p = proportion of adults who totally abstain from alcohol.</em></u>

where, p = \frac{429}{1100} = 0.39

So, Null Hypothesis, H_0 : p = 39%      {means that the proportion of adults who totally abstain from alcohol​ has not changed}

Alternate Hypothesis, H_A : p \neq 39%      {means that the proportion of adults who totally abstain from alcohol​ has changed}

The test statistics that would be used here <u>One-sample z proportion 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 adults who totally abstain from alcohol = \frac{352}{1100} = 0.32

           n = sample of adults surveyed = 1100

So, <u><em>test statistics</em></u>  =  \frac{0.32-0.39}{\sqrt{\frac{0.32(1-0.32)}{1100} } }  

                              =  -4.976

The value of z test statistics is -4.976.

<em>Now, at 0.10 significance level the z table gives critical values of -1.645 and 1.645 for two-tailed test.</em><em> </em>

<em>Since our test statistics doesn't lie 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 </em><em><u>we reject our null hypothesis</u></em><em>.</em>

Therefore, we conclude that the proportion of adults who totally abstain from alcohol​ has changed.

4 0
3 years ago
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