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vfiekz [6]
3 years ago
14

Please Hurry!!! THANKS!!

Mathematics
2 answers:
Phoenix [80]3 years ago
5 0

Answer:

first one is A. I have no idea what the second is. good luck, it might be D.

antoniya [11.8K]3 years ago
4 0

Answer:

1) 3x³ - x² - 15x + 5

2) x² - 4

3) (5, 3)

4) f⁻¹(x) = 1/3x - 4

Step-by-step explanation:

1......................

<h3>Given</h3>
  • f(x) = 3x - 1
  • g(x) = x² - 5
  • f(x) · g(x) =?
<h3>Solution</h3>
  • f(x) · g(x) = (3x - 1)(x² - 5) =
  • 3x·x² - 3x·5 - x² + 5 =
  • 3x³ - x² - 15x + 5

Correct option is A.

2......................

<h3>Given</h3>
  • f(x) = x²
  • g(x) = x - 4
  • g(f(x))  =?
<h3>Solution</h3>
  • g(f(x)) = g(x²) = x² - 4

Correct option is D.

3......................

Point (5, 3) is the inverse of point (3,5)

Correct option is C

4......................

<h3>Given</h3>
  • f(x) = 3x - 12
  • Inverse of function = ?
<h3>Solution</h3>

To get the inverse of the function, switch over domain and range:

  • x = 3f⁻¹(x) - 12
  • 3f⁻¹(x) = x - 12
  • f⁻¹(x) = 1/3x - 4

Correct option is D.

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Alik [6]

Answer:

z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575  

p_v =2*P(z>1.575)=0.115  

So the p value obtained was a very high value and using the significance level given \alpha=0.01 we have 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 businesses were highly profitable is not significantly different from 0.28 or 28%.

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

X=19 represent the businesses were highly profitable

\hat p=\frac{19}{50}=0.38 estimated proportion of businesses were highly profitable

p_o=0.28 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of businesses were highly profitable is different from 0.28 or no, the system of hypothesis is.:  

Null hypothesis:p=0.28  

Alternative hypothesis:p \neq 0.28  

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 required we can replace in formula (1) like this:  

z=\frac{0.38 -0.28}{\sqrt{\frac{0.28(1-0.28)}{50}}}=1.575  

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=0.01. 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>1.575)=0.115  

So the p value obtained was a very high value and using the significance level given \alpha=0.01 we have 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 businesses were highly profitable is not significantly different from 0.28 or 28%.

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