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Sav [38]
2 years ago
8

I need help with finding the equation for #20 and # 22 for my brother we get the answer but it’s hard for us to write out how to

solve them both by showing the work which if He doesn’t show the work then won’t receive full credit! Please help he’s in 4th grade

Mathematics
1 answer:
Pavel [41]2 years ago
8 0

Answer:

Im not sure about 20 but 22 is 48

Step-by-step explanation:

4×96 =384

384÷8 = 48 orange per classroom

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Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

Step-by-step explanation:

We have the following info given:

Confidence= 0.95 the confidence level desired

ME =0.03 represent the margin of error desired

The margin of error for the proportion interval is given by this formula:  

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

The confidence level is 95% or 0.95, the significance is \alpha=0.05 and the critical value for this case using the normal standard distribution would be z_{\alpha/2}=1.96

Since we don't have prior information we can use \hat p= 0.5 as an unbiased estimator

Also we know that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

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1 year ago
What adds to negative 8 and multiplies to negative 10
butalik [34]
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