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SVETLANKA909090 [29]
3 years ago
7

Drag the expressions to the correct functions. Not all expressions will be used. Consider the functions f and g. f(x)=4x^2+1 g(x

)=x^2-3 Perform the function compositions.

Mathematics
1 answer:
bagirrra123 [75]3 years ago
5 0

Answer:

(f o g)(x) = 4x⁴ - 96x + 37

(g o f)(x) = 16x⁴ + 8x² - 2

Step-by-step explanation:

Since, the given functions are,

f(x) = 4x² + 1

g(x) = x² - 3

(f o g)(x) = f[g(x)]

f[g(x)] = 4(x² - 3)² + 1

         = 4(x⁴ - 24x + 9) + 1

         = 4x⁴ - 96x + 36 + 1

         = 4x⁴ - 96x + 37

(g o f)(x) = g[f(x)]

             = (4x² + 1)²- 3

             = 16x⁴ + 8x² + 1 - 3

             = 16x⁴ + 8x² - 2

Therefore, out of the given options only two expressions are correct.              

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Last year, a comprehensive report stated that 28% of businesses in the northeast of Ohio were considered highly profitable. This
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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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