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Natalija [7]
2 years ago
12

A survey of 500 adults aged 18 – 29 years of age revealed that 285 chose to eat fast food for dinner at least once in the past w

eek. Find the sample proportion of individuals surveyed who ate fast food for dinner at least once in the past week.
Mathematics
1 answer:
irina [24]2 years ago
3 0

Answer:

The sample proportion of individuals surveyed who ate fast food for dinner at least once in the past week is 0.57

Step-by-step explanation:

The sample proportion of individuals surveyed who ate fast food for dinner at least once in the past week is:

The number of people who ate fast food for dinner at least once in the past week divided by the size of the sample.

In this question:

Sample of 500 adults.

Of those, 285 chose to eat fast food for dinner at least once in the past week.

285/500 = 0.57

The sample proportion of individuals surveyed who ate fast food for dinner at least once in the past week is 0.57

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Answer:

h = 27.3

Step-by-step explanation:

A = bh

h = A/b

h = 210.21 / 7.7

h = 27.3

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Using parity, the probability for detecting an error, given that one has occurred, is:a. about 50% for either even or odd parity
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A student bought a truck with a down payment of 4000 and monthly payments of $250 for four years. What was the total cost of the
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4000+250(12)(4)

Down Payment plus the price per month, multiplied by the years. The total cost should be $16000.
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2 years ago
A sample of 1714 cultures from individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiot
Damm [24]

Answer:

a) p represent the real population proportion of people who showed partial or complete resistance to the antibiotic

b) \hat p=\frac{973}{1714}=0.568 represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

c) We are confident (95%) that the true proportion of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic is between 54.5% to 59.1%.  

Step-by-step explanation:

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".

Part a

Description in words of the parameter p

p represent the real population proportion of people who showed partial or complete resistance to the antibiotic

\hat p represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

z_{\alpha/2} represent the critical value for the margin of error

The population proportion have the following distribution  

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

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{973}{1714}=0.568 represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is 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

And replacing into the confidence interval formula we got:

0.568 - 1.96 \sqrt{\frac{0.568(1-0.568)}{1714}}=0.545

0.568 + 1.96 \sqrt{\frac{0.56(1-0.56)}{1714}}=0.591

And the 95% confidence interval would be given (0.545;0.591).

We are confident that about 54.5% to 59.1% of individuals in Florida diagnosed with a strep infection was tested for resistance to the antibiotic penicillin present partial or complete resistance to the antibiotic.  

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2 years ago
7.Find a quadratic polynomial whose zeros are 5 and -5
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sorry i dont understand the question

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