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Nata [24]
3 years ago
8

The dimensions of a rectangle can be represented by the functions shown. Which function represents the area of the rectangle, h(

x) = f(x)g(x)? h(x) = 15x2 – 29x 14 h(x) = 15x2 – 41x – 9 h(x) = 15x2 – 29x – 14 h(x) = 15x2 – 41x 14.
Mathematics
1 answer:
marta [7]3 years ago
5 0

The area of the rectangle is represented by the function h(x) = 15x^2 - 41x + 14.

Option D is the correct answer.

<h3>How do you calculate the area of the rectangle?</h3>

Given that the functions are,

f(x) = 5x-2 and g(x)= 3x-7

The area of the rectangle is given below.

h(x) = f(x)g(x)

h(x) = (5x-2)(3x-7)

h(x) = 15x^2 - 35x - 6x + 14

h(x) = 15x^2 - 41x + 14

Hence the area of the rectangle is represented by the function h(x) = 15x^2 - 41x + 14. Option D is the correct answer.

To know more about the area of the rectangle, follow the link given below.

brainly.com/question/16719731.

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Answer:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

Step-by-step explanation:

Information given

data: 202.2 203.4 200.5 202.5 206.3 198.0 203.7 200.8 201.3 199.0

We can calculate the sample mean and deviation with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X=201.77 represent the sample mean    

s=2.41 represent the sample standard deviation    

n=10 sample size    

\mu_o =200 represent the value that we want to test    

\alpha=0.05 represent the significance level for the hypothesis test.    

t would represent the statistic

p_v represent the p value for the test

Hypothesis to test

We want to determine if the true mean is equal to 200, the system of hypothesis are :    

Null hypothesis:\mu = 200    

Alternative hypothesis:\mu = 200    

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)    

The statistic is given by:

t=\frac{201.77-200}{\frac{2.41}{\sqrt{10}}}=2.32    

The degrees of freedom are given by:

df=n-1=10-1=9  

The p value for this case is given by:

p_v =2*P(t_{(9)}>2.32)=0.0455    

For this case since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly different from 200 F.

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Answer:

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Step-by-step explanation:

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