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1 year ago

How do I find the domain range and function of the graph

Mathematics

1 answer:

Allisa [31]1 year ago

3 0

Answer:

D = {-3, -2, -1, 1, 4}

R = {-5, -1, 0, 2}

Step-by-step explanation:

You have correctly described the process of finding the domain and range by the way you filled in the blanks at the top of the sheet.

The domain is the set of x-values of the points on the graph:

D = {-3, -2, -1, 1, 4}

The range is the set of (unique) y-values of the points on the graph:

R = {-5, -1, 0, 2}

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Vitamin D, whether ingested as a dietary supplement or produced naturally when sunlight falls on the skin, is essential for stro

tigry1 [53]

Answer:

a) The 98% confidence interval would be given (0.182;0.218).

b) We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

c) If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

d) Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

Step-by-step explanation:

Data given and notation

n=2700 represent the random sample taken

X represent the people in England who are deficient in Vitamin D

$\hat p=0.2$ estimated proportion of England people who are deficient in Vitamin D

$\alpha=0.02$ represent the significance level

Confidence =0.98 or 98%

z would represent the statistic (variable of interest)

p= population proportion of England people who are deficient in Vitamin D

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 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

And replacing into the confidence interval formula we got:

$0.20 - 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.182$

$0.20 + 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.218$

And the 98% confidence interval would be given (0.182;0.218).

Part b

We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

Part c

If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

Part d

Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

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2 years ago

you buy dog biscuits at the bulk rate of 2.25/lb. the scale shows .75 lbs fir 25 biscuits. find cost of 40 biscuits

julia-pushkina [17]

1.2 pounds is the weight of 40 biscuits and the cost is $2.70.

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3 years ago

Read 2 more answers

The principal of a large high school wants to know if students spend more than 1 hour doing homework per night, on average. To i

MrRissso [65]

Answer:

If we define the random variable X ="time spend by the students doign homework"

And we want to tes t is students spend more than 1 hour doing homework per night, on average (alternative hypothesis), so then the system of hypothesis for this case are:

Null hypothesis: $\mu \leq 1$

Alternative hypothesis: $\mu >1$

And they wnat to use a sample size of n = 100 and a significance level of 0.05

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

Solution to the problem

If we define the random variable X ="time spend by the students doign homework"

And we want to tes t is students spend more than 1 hour doing homework per night, on average (alternative hypothesis), so then the system of hypothesis for this case are:

Null hypothesis: $\mu \leq 1$

Alternative hypothesis: $\mu >1$

And they wnat to use a sample size of n = 100 and a significance level of 0.05

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3 years ago

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Mariulka [41]

Help with what? I am not sure what other question you asked that you need help on. What is the question?

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Answer:

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Step-by-step explanation:

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