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BigorU [14]
3 years ago
10

B. Solve;

Mathematics
1 answer:
Contact [7]3 years ago
3 0
1 .
Let’s say your number is x.
3x+5>33
3x>28
x>28/3
2.
Let’s say the two consecutive integers are x and x-1
Since the two numbers are integers and two consecutive numbers (odd+even) can never add up to an even number (80 for example).
The highest possible sum of two consecutive integers that is less than 81 will add up to 79.

x+x-1=79
2x-1=79
2x=80
x=40
So the two consecutive numbers are 40 (x) and 39(x-1)
Hope this helps


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For this case the margin of error for a proportion is given by this formula

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

For this case the confidence level is 90% or 0.9 so then the significance would be

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With \alpha/2=0.05, we can find the value for z_{\alpha/2} using the normal standard distribution table or excel.

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

Squaring both sides:

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

And solving for n we got:

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

Now we can replpace the values

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.644854})^2}=422.739

And rounded up to the nearest integer we got:

n=423

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