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sp2606 [1]
3 years ago
14

6 weeks = 6*(1 week)=6* (7 days)= 42 days​

Mathematics
2 answers:
balu736 [363]3 years ago
7 0

Answer:

42

Step-by-step explanation:

Yes, you are right.

6 weeks * 7 days in a week.

so 7 days in all 6 weeks is 42 days.

Delicious77 [7]3 years ago
6 0
You are correct. If that is what you are asking.
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What percent of 156 is 78
alina1380 [7]
50% is your answer hope i helped you
4 0
3 years ago
The attendance at a family reunion was 160 people. This was 125% of last year's attendance. How many people attended the reunion
Alex787 [66]
If you would like to know how many people attended the reunion last year, you can calculate this using the following steps:

125% of last year's attendance is 160 people
125% of x is 160
125% * x = 160
125/100 * x = 160     /*100/125
x = 160 * 100 / 125
x = 128 people (last year)

Result: 128 people attended the reunion last year.
4 0
3 years ago
SOMEONE PLEASE HELP ME I DONT KNOW HOW TO DO THIS AT ALL
Lina20 [59]

Answer:

y=2cos(4\pi x)\\


Step-by-step explanation:

Since the amplitude of the cosine function is 2, I would put the multiply it by 2. There is no phase shift

The period is 1/2, so it would be 4pi

Plug them in into y=Acos(Bx), y=2cos(4pix)

6 0
2 years ago
At a charity fundraiser, adult tickets are sold for $12 each and children tickets are sold for $6 each. Write an algebraic expre
Annette [7]

Answer:

the answer is 2

Step-by-step explanation:

7 0
3 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
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