What is 100% of 94 in a decimal

6(33 - w) = 90 solve for w entre your answer in the box w =

aliina [53]

Answer:

w = 18

Step-by-step explanation:

6(33 - w) = 90

33 - w = 15

-w = -18

w = 18

5 0

2 years ago

Solve for -6+3x=x-2+4x

Iteru [2.4K]

Answer:

x= -2

Step-by-step explanation: If you feel skeptical about my answer, then you should plug -2 back into the equation to check if it's correct. And it's correct for me when I did it.

7 0

3 years ago

Someone help me out pleasee

Vanyuwa [196]

Answer:

-23

Step-by-step explanation:

The discriminant is the number inside of the radical of the quadratic formula.

5 0

3 years ago

For brainlest pls ..

mixer [17]

39 (Hope you understand )

4 0

3 years ago

Read 2 more answers

A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in deter

lbvjy [14]

Answer:

$z=\frac{0.85 -0.8}{\sqrt{\frac{0.8(1-0.8)}{100}}}=1.25$

$p_v =P(z>1.25)=0.106$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who favored candidate A is not significalty higher than 0.8

ii) not significantly greater than 80%

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

$\hat p=0.85$ estimated proportion of people who favored candidate A

$p_o=0.8$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 0.8.:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

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:

$z=\frac{0.85 -0.8}{\sqrt{\frac{0.8(1-0.8)}{100}}}=1.25$

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 $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z>1.25)=0.106$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who favored candidate A is not significalty higher than 0.8

ii) not significantly greater than 80%

8 0

3 years ago