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Thepotemich [5.8K]
2 years ago
9

Can u show me how to solve this problem 0.11 times 0.91

Mathematics
1 answer:
lisov135 [29]2 years ago
3 0

Answer:

0.1001

Step-by-step explanation: Your answer is 0.1001

 0.91

 <u> 0.11</u>

   1011

 0990

<u>00000</u>

0.1001


Hope this helps you!!! :)

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Yes 8+3= 11 You regroup the one 4+2=6+1=7 Hope this helps!
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Match each operation involving f(x) and g(x) to its answer
eimsori [14]

Answer:

(g-f) (-1)= sqrt(15)

(f/g)(-1)= 0

(g+f)(2)=sqrt(3)-3

(g*f)(2)=-3*sqrt(3)

Step-by-step explanation:

We have to eval the expressions given in the point indicated.

Lets start by the first equation

(g-f)(-1)= g(-1) - f(-1)= \sqrt{11-4*(-1)}    - 1 +(-1)^{2} = \sqrt{15}

Now, lest continue with the others

(f/g)(-1)= f(-1)/g(-1)= (1-1)/sqrt(15)=0

(g+f)(2)=g(2)+f(2)=sqrt(3)-3

(g*f)(2)=g(2)*f(2)=sqrt(3)*(-3)=-3sqrt(3)

3 0
3 years ago
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7 0
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A private and a public university are located in the same city. For the private university, 1038 alumni were surveyed and 647 sa
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Answer:

The difference in the sample proportions is not statistically significant at 0.05 significance level.

Step-by-step explanation:

Significance level is missing, it is  α=0.05

Let p(public) be the proportion of alumni of the public university who attended at least one class reunion  

p(private) be the proportion of alumni of the private university who attended at least one class reunion  

Hypotheses are:

H_{0}: p(public) = p(private)

H_{a}: p(public) ≠ p(private)

The formula for the test statistic is given as:

z=\frac{p1-p2}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}} where

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  • p2 is the sample proportion of private university students who attended at least one class reunion  (\frac{647}{1038}=0.623)
  • p is the pool proportion of p1 and p2 (\frac{808+647}{1311+1038}=0.619)
  • n1 is the sample size of the alumni from public university (1311)
  • n2 is the sample size of the students from private university (1038)

Then z=\frac{0.616-0.623}{\sqrt{{0.619*0.381*(\frac{1}{1311} +\frac{1}{1038}) }}} =-0.207

Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.  

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