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ratelena [41]
3 years ago
10

Dr potter provides vaccinations against polio and measles. Each polio vaccination multi-dose vial consists of 44 individual dose

s, and each measles vaccination multi-dose consists of 22 individual doses. Last year, Dr. Potter used a total of 60 multi-dose vials that consisted of a total of 2024 individual doses. How many individual polio and measles vaccinations did Dr. Potter give respectively
Answer choices
A (28,32)
B(32,28)
C(64,32)
D(14,28)
Mathematics
1 answer:
creativ13 [48]3 years ago
5 0
The correct answer for the question that is being presented above is this one: "A (28,32)." Dr potter provides vaccinations against polio and measles. Each polio vaccination multi-dose vial consists of 44 individual doses, and each measles <span>vaccination multi-dose consists of 22 individual doses. Dr. Potter give 28 and 32 individual polio and measles vaccinations, respectively.</span>
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Of 575 broiler chickens purchased from various kinds of food stores in different regions of a country and tested for types of ba
chubhunter [2.5K]

Answer:

a) 0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

b) We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

c) For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

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".  

p represent the real population proportion of interest  

\hat p =0.68 represent the estimated proportion

n=575 is the sample size required (variable of interest)  

z represent the critical value for the margin of error  

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})  

Part a

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 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.  

z_{\alpha/2}=2.58  

And replacing into the confidence interval formula we got:  

0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630  

0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730  

And the 99% confidence interval would be given (0.630;0.730).  

Part b

We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

Part c

For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval  are satisfied, so then the results can be assumed for all the population

4 0
3 years ago
Refer to the triangle below. If a = 4 and b = 6, solve for c. show work Also, give the unsimplified radical form answer and then
9966 [12]

Look at the image below where I labeled the sides

To solve this you must use Pythagorean theorem:

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5 0
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How to Factor r^2-5r-14
12345 [234]

Answer:

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Step-by-step explanation:

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Now, let's look at the -5. How can we make -5 with the pairs that we just found?

The only possible way is 2 and -7 (added together equals -5)

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The cost of textbooks for a school increases with the average class size. Identify the independent and dependent quantity in the
Pani-rosa [81]
Cost depends on the class size  , so 
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Answer:

m = (5n/4) -7

Step-by-step explanation:

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