2a+ 4b +30<br><br><br> 16 points

Can someone clearly explain and workout Q19? Much appreciated

Anna007 [38]

The x coefficient is 2
Take half of 2 to get 2/2 = 1
Square 1 to get 1^2 = 1

Add and subtract 1 to the expression. This is the same as adding 0.

x^2 + 2x + 2
x^2 + 2x + 2 + 0
x^2 + 2x + 2 + (1-1)
(x^2 + 2x + 1) + 2 - 1
(x+1)^2 + 1

Answer: (x+1)^2 + 1

5 0

3 years ago

According to a Washington Post-ABC News poll, 331 of 502 randomly selected U.S. adults interviewed said they would not be bother

k0ka [10]

Answer:

$z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124$

$p_v =P(z>7.124)=5.24x10^{-13}$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is significantly higher than 0.5.

Step-by-step explanation:

Data given and notation

n=502 represent the random sample taken

X=331 represent the adults that said they would not be bothered if the NAtional security agency

$\hat p=\frac{331}{502}=0.659$ estimated proportion of people who would not be bothered if the NAtional security agency

$p_o=0.5$ 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 the majority of 0.5:

Null hypothesis: $p\leq 0.5$

Alternative hypothesis: $p > 0.5$

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:

$z=\frac{0.659 -0.5}{\sqrt{\frac{0.5(1-0.5)}{502}}}=7.124$

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:

$p_v =P(z>7.124)=5.24x10^{-13}$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is significantly higher than 0.5.

4 0

3 years ago

The decimal 0.4444 is an example of what kind of number

Slav-nsk [51]

That would be a decimal!

5 0

2 years ago

Evaluate the radical.

zaharov [31]

5*5*5=125
5²*5=125
5√5 is your answer

4 0

3 years ago