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liq [111]
4 years ago
6

Divide 4p2 + 6 by 2p – 2. Which statements are true about the division?

Mathematics
2 answers:
Goryan [66]4 years ago
8 0
The answer will be
2p+-3
podryga [215]4 years ago
7 0

Answer:

2p + -3

Step-by-step explanation:

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Suppose 244 subjects are treated with a drug that is used to treat pain and 51 of them developed nausea. Use a 0.01 significance
vodka [1.7K]

Answer:

z=\frac{0.209 -0.2}{\sqrt{\frac{0.2(1-0.2)}{244}}}=0.351  

p_v =P(z>0.351)=0.363  

So the p value obtained was a very high value and using the significance level given \alpha=0.01 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of subjects with nauseas is not higher than 0.2 or 20% at 1% of significance.

Step-by-step explanation:

Data given and notation  

n=244 represent the random sample taken

X=51 represent the subjects with nausea

\hat p=\frac{51}{244}=0.209 estimated proportion of subjects with nausea

p_o=0.2 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.2.:  

Null hypothesis:p \leq 0.2  

Alternative hypothesis:p > 0.2  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.209 -0.2}{\sqrt{\frac{0.2(1-0.2)}{244}}}=0.351  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(z>0.351)=0.363  

So the p value obtained was a very high value and using the significance level given \alpha=0.01 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the proportion of subjects with nauseas is not higher than 0.2 or 20% at 1% of significance.

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3 years ago
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18 = 2x +4

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14 = 2x

divide both sides by 2:

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Hi there :-)
Use the formula of the present value of annuity ordinary
Pv=pmt [(1-(1+r/k)^(-kn))÷(r/k)]
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K compounded monthly 12
N time 30 years

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