Solve this equation 4y-y+11=38

Maya's standard deduction on her federal income tax return is $8350. If she paid $5980 in state taxes and $2430 in mortgage inte

Aleonysh [2.5K]

B. No, because it's less than the deduction she would get from itemizing.-APEX

5 0

3 years ago

Read 2 more answers

Negative three times the sum of two times a number and 1 is equal to 5.” Write an equation that represents the equation, then so

Kamila [148]

So, we need to consider what is being done.

So, lets start with the -3.

-3*

Now, lets see what's next

2 times a number is the same as saying 2x, then add one.

That is equal to five.

-3(-2x+1)=5

Now,

6x-3 = 5

6x = 8

x= 8/6

8 0

3 years ago

Mark talked on his cell phone 3 hours over the weekend. Generic talked on his phone 1 9/10 hours. How much longer did mark talk

lubasha [3.4K]

Do 3 minus 1.9

- To make this easier I would do

3 - 2 = 1
1 + 0.1 = 1.1

8 0

3 years ago

Simplify the following expression. 3 2/5x3(-7/5)

Studentka2010 [4]

The simplified expression of 32/5 x 3(-7/5) is -672/25

<h3>How to simplify the expression?</h3>

The mathematical expression is given as:

32/5 x 3(-7/5)

Evaluate the product of 3 and -7/5

32/5 x 3(-7/5) = 32/5 x -21/5

Evaluate the product of 32 and -21

32/5 x 3(-7/5) = 672/5 x -1/5

Evaluate the product of 5 and 5

32/5 x 3(-7/5) = 672/25 x -1

So, we have:

32/5 x 3(-7/5) = -672/25

Hence, the simplified expression of 32/5 x 3(-7/5) is -672/25

Read more about expressions at:

brainly.com/question/723406

#SPJ1

8 0

2 years ago

In a study of 420,095 Danish cell phone users, 135 subjects developed cancer of the brain or nervous system (based on data from

Maksim231197 [3]

Answer:

$z=\frac{0.0003214 -0.00034}{\sqrt{\frac{0.00034(1-0.00034)}{420095}}}=-0.654$

$p_v =2*P(Z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.005$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that the true proportion not differs significantly from the specified value of 0.00034 or 0.034%.

Step-by-step explanation:

1) Data given and notation

n=420095 represent the random sample taken

X=135 represent the subjects developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute as reported in USA Today)

$\hat p=\frac{135}{420095}=0.0003214$ estimated proportion of subjects developed cancer of the brain or nervous system (based on data from the Journal of the National Cancer Institute as reported in USA Today)

$p_o=0.00034$ is the value that we want to test

$\alpha=0.005$ represent the significance level

Confidence=99.5% or 0.995

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 the brain or nervous system at a rate that is different from the rate of 0.0340% :

Null hypothesis: $p=0.00034$

Alternative hypothesis: $p \neq 0.00034$

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:

$z=\frac{0.0003214 -0.00034}{\sqrt{\frac{0.00034(1-0.00034)}{420095}}}=-0.654$

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 $\alpha=0.005$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(Z$

So the p value obtained was a very high value and using the significance level given $\alpha=0.005$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say that the true proportion not differs significantly from the specified value of 0.00034 or 0.034%.

4 0

3 years ago