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tiny-mole [99]
3 years ago
9

I’m Really lost if I could get a answer it would be greatly appreciated

Mathematics
1 answer:
Nataly [62]3 years ago
6 0
<h3>Answers:</h3>

f(g(x)) = \sqrt{x^2+5}+5\\\\g(f(x)) = x+30+10\sqrt{x-1}

================================================

Work Shown:

Part 1

f(x) = \sqrt{x-1}+5\\\\f(g(x)) = \sqrt{g(x)-1}+5\\\\f(g(x)) = \sqrt{x^2+6-1}+5\\\\f(g(x)) = \sqrt{x^2+5}+5\\\\

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.

------------------

Part 2

g(x) = x^2+6\\\\g(f(x)) = \left(f(x)\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}+5\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}\right)^2+2*5*\sqrt{x-1}+\left(5\right)^2+6\\\\g(f(x)) = x-1+10\sqrt{x-1}+25+6\\\\g(f(x)) = x+30+10\sqrt{x-1}\\\\

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2

In this case, a = sqrt(x-1) and b = 5.

You could also use the box method as a visual way to expand out \left(\sqrt{x-1}+5\right)^2

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Answer:

z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32  

p_v =P(z  

Since the p value is lower than the significance level \alpha=0.1 we have enough evidence to reject the null hypothesis and the claim for the manager makes sense.

For the Ti84 preocedure we need to do this:

STAT> TESTS> 1-Z prop Test

And then we need to input the following values:

po= 0.58

x = 443 , n= 820

prop <po

And then calculate and we will get the same results

Step-by-step explanation:

Data given and notation

n=820 represent the random sample taken

X=443 represent the people that had cell phones

\hat p=\frac{443}{820}=0.540 estimated proportion of people that had cell phones

p_o=0.58 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

System of hypothesis

We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.58.:  

Null hypothesis:p\geq 0.58  

Alternative hypothesis:p < 0.58  

The statistic is given by:

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

And replacing we got:

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

z=\frac{0.54 -0.58}{\sqrt{\frac{0.58(1-0.58)}{820}}}=-2.32  

Decision

The significance level provided \alpha=0.1. Now we can calculate the p value

Since is a left tailed test the p value would be:  

p_v =P(z  

Since the p value is lower than the significance level \alpha=0.1 we have enough evidence to reject the null hypothesis and the claim for the manager makes sense.

For the Ti84 preocedure we need to do this:

STAT> TESTS> 1-Z prop Test

And then we need to input the following values:

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