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3 years ago

Will Mark Brainlest help me please

Mathematics

1 answer:

Serhud [2]3 years ago

5 0

p(x) × q(x)

= (3x² - x + 1) × (2x² + 3x - 2)

= 6x⁴ + 9x³ - 6x² - 2x³ - 3x² + 2x + 2x² + 3x - 2

= 6x⁴ + 7x³ - 7x² + 5x - 2

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Step-by-step explanation:

So, 6 feet is equal to 72 inches. You need 18 inches for one bracelet, so by dividing 72/18, you get 4. So, he can make 4 bracelets.

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Does going to a private university increase the chance that a student will graduate with student loan debt? A national poll by t

Snowcat [4.5K]

Answer:

Null hypothesis: $p \leq 0.69$

Alternative hypothesis: $p > 0.69$

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the number of graduates that had student loan debt in 2014

$\hat p=0.71$ estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

$p_o=0.69$ is the value that we want to test

$\alpha$ represent the significance level

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 if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014 respect to the value of 2013.:

Null hypothesis: $p \leq 0.69$

Alternative hypothesis: $p > 0.69$

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:

$z=\frac{0.71 -0.69}{\sqrt{\frac{0.69(1-0.69)}{1500}}}=1.675$

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 assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(Z>1.675)=0.047$

If we compare the p value obtained and using the significance level assumed $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults that said that it is morally wrong to not report all income on tax returns is not significantly higher than 0.69.