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Troyanec [42]
3 years ago
9

At the luau, a vender was selling 5 leis for $25. what is the unit price?

Mathematics
1 answer:
gayaneshka [121]3 years ago
5 0

Answer:

$5

Step-by-step explanation:

if 5 leis are worth $25, each lei would be 25/5

25/5 is 5/1, when the denominator is one, that is unit price

the unit price is $5

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When the 2010 General Social Survey asked subjects in the US whether they would be willing to accept cuts in their standard of l
Triss [41]

Answer:

a) \hat p=\frac{486}{1374}=0.354 estimated proportion of adults that they would be willing to accept cuts in their standard of living to protect the environment

b) 0.354 - 2.58 \sqrt{\frac{0.354(1-0.354)}{1374}}=0.321

0.354 + 2.58 \sqrt{\frac{0.354(1-0.354)}{1374}}=0.387

And the 90% confidence interval would be given (0.321;0.387).

c) z=\frac{0.354 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1374}}}=-10.82  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportion of the population that would say yes is significantly lower than 0.5.

Step-by-step explanation:

Data given and notation

n=1374 represent the random sample taken

X=486 represent the adults that said that they would be willing to accept cuts in their standard of living to protect the environment

Part a

\hat p=\frac{486}{1374}=0.354 estimated proportion of adults that they would be willing to accept cuts in their standard of living to protect the environment

Part b

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:

0.354 - 2.58 \sqrt{\frac{0.354(1-0.354)}{1374}}=0.321

0.354 + 2.58 \sqrt{\frac{0.354(1-0.354)}{1374}}=0.387

And the 90% confidence interval would be given (0.321;0.387).

Part c

p_o=0.5 is the value that we want to test

\alpha=0.01 represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

We need to conduct a hypothesis in order to test the claim that whether a majority or minority of the population would say yes:  

Null hypothesis:p\geq 0.5  

Alternative hypothesis:p < 0.5  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.354 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1374}}}=-10.82  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is \alpha=0.01. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.01 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportion of the population that would say yes is significantly lower than 0.5.  

8 0
3 years ago
Suppose an electric-vehicle manufacturing company estimates that a driver who commutes 50 miles per day in a particular vehicle
Mariana [72]

Using the uniform distribution, it is found that the mathematical expression for the probability density function of battery recharging time for this scenario is given by:

f(x) = \frac{1}{40}, 70 \leq x \leq 110

f(x) = 0, elsewhere.

<h3>What is the uniform probability distribution?</h3>

It is a distribution with two bounds, a and b, in which each outcome is equally as likely.

The probability density function is:

f(x) = \frac{1}{b - a}, a \leq x \leq b

f(x) = 0, elsewhere.

In this problem, the actual recharging time required is uniformly distributed between 70 and 110 minutes, hence a = 70, b = 110, and the density function is:

f(x) = \frac{1}{40}, 70 \leq x \leq 110

f(x) = 0, elsewhere.

More can be learned about the uniform distribution at brainly.com/question/13889040

6 0
2 years ago
Evaluate the expression shown below and write your answer as a fraction in simplest form. 16/11 − 12/1
Zepler [3.9K]

Answer:

-10 6/11

Step-by-step explanation:

16/11−12/1=16/11+−12/1=16/11+−132/11=16+−132/11=−116/11


-116/11 = −10 6/11

7 0
2 years ago
Cash converters bought tv for 1500.if the repair cost them 347 and they resold it for 2500,what percentage profit did they make
DerKrebs [107]

Answer:

good Luck!

Follow me!

<u>david_201822</u>

3 0
3 years ago
Lets find 1/3 - 1/12<br><br> 1/3 - 1/12 = ___/12 - 1/12 =___/___<br><br> Fill in the blanks
kicyunya [14]
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