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KonstantinChe [14]
3 years ago
8

Does anyone know how to do equations in 6th-grade math? i will be rewarding 15 points to whoever helps me

Mathematics
2 answers:
bazaltina [42]3 years ago
8 0
Yea I can help you with these problems I’ve done this a long time ago and so far I remembered how to do this. So for number 1. X=4 in this problem so we need to substitute x into 4 and recreate this expression: 4+1=5, and yes it is a solution. As for number 2, w would equal to 2 so 13-2=10 isn’t a solution because 13-12 is actually 11, so this is a false. Now number 3, v would equal to 10 so you’d have to replace 10 into this expression; so it’s going to be 2(10)=12 which isn’t a solution because 2 times 10 equals to 20. Now for number 4, p would have to equal to 7, here is the expression; 14/7=2, is a solution because if you’re going to check it do 7(2) and it would actually equal to 14 so yes it is a solution. Now for number 5 you would have to equal w to 3 so do this in an expression; 8+3=11, and yes that is a solution because you can check it as 11-3 which would equal to 8. Now for number 6 t would have to equal to 5 so replace t into 5 which the expression is going to be; 4(5)=20 so yes it is a solution because you can check it by doing 20/5=4. Now for number 7 s would have to equal to 3 so in this expression would have to be; 12/3=4 and yes that is a solution because you can check it by doing 3 times 4 which would equal to 12. Now for the last problem 8, d would have to equal to 8 so you replace d into 8 which the expression is going to be; 6+8=14, it’s not going to be 6+8=15 because it doesn’t equal to that since if you want to check the right answer just subtract 14 by 8 and you would get 6. So the last problem isn’t going to be a solution.

Ann [662]3 years ago
6 0

Answer:

yeah sure

Step-by-step explanation:

the answer is there already right

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4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B
AleksAgata [21]

Answer:

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

p_v =P(Z>2.123)=0.0169  

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 lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

\hat p=\frac{63}{200}=0.315 estimated proportion of lawyers had used some form of advertising for their business

p_o=0.25 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)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

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.

3) Calculate the statistic  

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

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

4) 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>2.123)=0.0169  

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 lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

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Answer:

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Step-by-step explanation:

48 = 12 * 4

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