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lisov135 [29]
2 years ago
11

Ric is half Paul's age. Tom is 2 years younger than Paul. The sum of all their ages is 33 years Show all working clearly to find

Ric's age​
Mathematics
1 answer:
VMariaS [17]2 years ago
4 0

Answer:

Ric is 7 years old.

Step-by-step explanation:

Choose variables for everyone's ages. I'll choose r for Ric's age, p for Paul's age, and t for Tom's age.

Since Ric's age is half of Paul's age, we can write this equation:

r=\frac{1}{2}p

Since Tom is two years younger than Paul, we can write this equation:

t=p-2

Since the sum of all their ages is 33 years, we can write this equation:

r+p+t=33

Since we know t and r in terms of p, we can replace them in the equation above as follows:

r+p+t=33\\(\frac{1}{2}p)+p+(p-2)=33

Now we can solve for p to find r later:

(\frac{1}{2}p)+p+(p-2)=33\\\frac{5}{2}p-2=33\\\frac{5}{2}p=35\\5p=70\\p=14

Now that we know that p=14, we can substitute 14 for p\\ in the equationr=\frac{1}{2}p:

r=\frac{1}{2}p\\r=\frac{1}{2}(14)\\r=7

Ric is 7 years old.

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Step-by-step explanation:

Data given and notation  

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\hat p=0.52 estimated proportion of of U.S. employers were likely to require higher employee contributions for health care coverage

\alpha=0.05 represent the significance level (no given, but is assumed)    

Solution to the problem

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The margin of error for this case is given by:

ME= z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

And replacing we got:

ME = 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.0259

And replacing into the confidence interval formula we got:

0.52 - 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.4941

0.52 + 1.64*\sqrt{\frac{0.52(1-0.52)}{1000}}=0.5459

And the 95% confidence interval would be given (0.4941;0.5459).

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