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mylen [45]
2 years ago
8

Help me with this question please!

Mathematics
1 answer:
BlackZzzverrR [31]2 years ago
7 0

\large \rm  \frac{2 ^{3} }{  \sqrt[4]{2}  }

\large \rm  \frac{2 ^{3} }{2 ^{ \frac{1}{4} } }

  • Divide the terms with the same base by subtracting their exponents

\large \rm \: 2 ^{3 -  \frac{1}{4} }

  • Calculate the difference

\large \rm2 ^{ \frac{11}{4} }

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Which shows another way to write 64?
Taya2010 [7]

None of those show another way to 64

6*6*6*6=1296

4*4*4*4*4*4=4096

6+6+6+6=24

6*4=24

Another way to show 64 is 8*8 or 4^3

4 0
4 years ago
At a​ school, 174 students play at least one sport. This is 75​% of the students at the school. How many students are at the​ sc
boyakko [2]

Answer:

  There are 232 students at the school.

Step-by-step explanation:

Let s represent the number of students at the school. We are told ...

  174 = 75% × s

  174/0.75 = s = 232 . . . . . divide by the coefficient of s

There are 232 students at the school.

3 0
3 years ago
Your uncle is currently four times as old as you are. In five years, your uncle will be three times as old as you. What is your
sertanlavr [38]

Answer:

your uncle will always be 4 times as old as you if he is currently 4 times as old as you

Step-by-step explanation:

kinda self explanatory so i think its a trick question haha

8 0
3 years ago
I need help pleasee!!!!
andre [41]

Answer:

Your answer is c

Step-by-step explanation:

the two inequalities are x =< 1 and x =>2 which is shown by the number line in option c

8 0
3 years ago
in a program designed to help patients stop smoking 232 patients were given sustained care and 84.9% of them were no longer smok
grandymaker [24]

Answer:

z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869  

p_v =2*P(Z>1.869)=0.0616  

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 adults were no longer smoking after one month is not significantly different from 0.8 or 80% .  

Step-by-step explanation:

1) Data given and notation

n=232 represent the random sample taken

X represent the adults were no longer smoking after one month

\hat p=0.849 estimated proportion of adults were no longer smoking after one month

p_o=0.80 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 the true proportion is 0.8.:  

Null hypothesis:p=0.8  

Alternative hypothesis:p \neq 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.

3) Calculate the statistic  

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

z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869  

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 bilateral test the p value would be:  

p_v =2*P(Z>1.869)=0.0616  

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 adults were no longer smoking after one month is not significantly different from 0.8 or 80% .  

6 0
3 years ago
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