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3 years ago

When x=3 and y=5 by how much does the value of 3x^2-2y exceed the value of 2x^2-3/

Mathematics

2 answers:

forsale [732]3 years ago

8 0

x = 3

y = 5

3x^2 – 2y

= 3(3)^2 - 2(5)

=> 3(9) - 10 = 17

And 2x^2– 3y

=> 2(3)^2 - 3(5)

=> 2(9) - 15 = 3

17 - 3 = 14

This gives 3x^2 – 2y exceeding 2x^2– 3y by 17 - 3 = 14

garri49 [273]3 years ago

4 0

14 is the correct answer.

When you use x = 3 and

y = 5 in the given expressions, 3x2 – 2y = 3(3)2– 2(5) = 27 – 10 = 17 and

2x2 – 3y = 2(3)2 – 3(5) = 18 – 15 = 3.

Then subtract 3 from 17....17-3 = 14.

14 is your answer.

Hope I helped ;]

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2 years ago

The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of m

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Answer:

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

$X \sim N(14.7,3.7)$

Where $\mu=14.7$ and $\sigma=3.7$

Since the distribution of X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we have;

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$\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59$

Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of $\bar X$ Round your answers to two decimal places.

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

$X \sim N(14.7,3.7)$

Where $\mu=14.7$ and $\sigma=3.7$

Since the distribution of X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we have;

$\mu_{\bar X}= 14.70$

$\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59$

4 0

3 years ago

in a program designed to help patients stop smoking 232 patients were given sustained care and 84.9% of them were no longer smok

grandymaker [24]

Answer:

$z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869$

$p_v =2*P(Z>1.869)=0.0616$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults were no longer smoking after one month is not significantly different from 0.8 or 80% .

Step-by-step explanation:

1) Data given and notation

n=232 represent the random sample taken

X represent the adults were no longer smoking after one month

$\hat p=0.849$ estimated proportion of adults were no longer smoking after one month

$p_o=0.80$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is 0.8.:

Null hypothesis: $p=0.8$

Alternative hypothesis: $p \neq 0.8$

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 $p_o$ .

3) Calculate the statistic

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

$z=\frac{0.849 -0.8}{\sqrt{\frac{0.8(1-0.8)}{232}}}=1.869$

4) 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 provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

$p_v =2*P(Z>1.869)=0.0616$

6 0

4 years ago