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tamaranim1 [39]
3 years ago
9

Help me out please need this

Mathematics
1 answer:
olchik [2.2K]3 years ago
5 0

the answer is ur guy

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Nitrates are groundwater contaminants derived from fertilizer, septic tank seepage and other sewage. Nitrate poisoning is partic
o-na [289]

Answer:

a) n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401  

And rounded up we have that n=2401

b) n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22  

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

Part a

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)  

And on this case we have that ME =\pm 0.02 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)  

Since we don't have prior information about the population proportion we can use ase estimator \hat p =0.5. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.02}{1.96})^2}=2401  

And rounded up we have that n=2401

Part b

For thi case the estimator for p is \hat p =0.07

n=\frac{0.07(1-0.07)}{(\frac{0.02}{1.96})^2}=625.22  

And rounded up we have that n=626

And we see that if we have previous info about p the sample size varies a lot.

4 0
3 years ago
A pet supply store sells a 20-pound bag of dog food $23.44. find the unit price what is the error of 20 divided by 23.44=.85
dimulka [17.4K]
23.44 / 20 = 1.172 rounds to 1.17 per lb <== this is cost per lb...this is telling u how much each lb of dog food costs

20 / 23.44 = 0.85....this is lbs per dollar....this is telling u that u r getting 0.85 lbs for a dollar
4 0
3 years ago
True or false cause I honestly forgot
fenix001 [56]

Answer:

false

Step-by-step explanation:

5 cant be put into 2

7 0
3 years ago
Read 2 more answers
Rewrite the function by completing the square<br><br> h(x)=4x^2+4x+1
lisov135 [29]

Answer:

h(x) = 4(x+\frac{1}{2})^{2}

Step-by-step explanation:

h(x)=4x^2+4x+1\\h(x)=4(x^2+x+\frac{1}{4}) \\h(x)=4(x^2+x+\frac{1}{4})\\h(x) = 4(x+\frac{1}{2})^{2}

8 0
2 years ago
1.6 liters of drink were equally distributed into 20 cups how many liters of drink would be in each cup
zimovet [89]

Answer:

0.08

Step-by-step explanation:

1.6 liters of the drink was distributed in 20 cups

The litres of drink in each cup can be calculated as follows

= 1.6/20

= 0.08

Hence 0.08 liters of drink would be in each cup

6 0
3 years ago
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