which function increases at a faster rate on 0 to infinity, f(x) = x2 or g(x) = 2x? explain your reasoning

Mathematics

2 answers:

Helga [31]2 years ago

8 0

Answer:

Using a table of values, the outputs of f(x) for whole numbers are 0, 1, 4, 9, 16, 25, 36, and so on. For the same input values, g(x) has outputs of 1, 2, 4, 8, 16, 32, and 64. Continuing to double the output each time results in larger outputs than those of f(x). The exponential function, g(x), has a constant multiplicative rate of change and will increase at a faster rate than the quadratic function.

Step-by-step explanation:

Tcecarenko [31]2 years ago

7 0

F(x) = x² increases at a faster rate than g(x) = 2x.

f(x)

Reason:

x = 0, 1, 2, 3, 4, 5, 6, 7

f(x) = 0, 1, 4, 9, 16, 25, 36, 49

g(x) = 0, 2, 4, 6, 8, 10, 12, 14.

Comparing the values of f(x) and g(x), we can see that that of f(x) are far higher than that of g(x) for the same values of x.

So f(x) increases at a faster rate.

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The equation yˆ=−6.2x2+51.5x+10.2 approximates the average number of cars that pass through an intersection x hours after 3:

Mars2501 [29]

Notice that form 3 pm to 6:30 pm 3.5 hours have passed.
Since the function $y=-6 x^{2} +51.5x+10.2$ represent the average number of cars that pass through an intersection x hours after 3:00 p.m, we are going to replace $x$ with 3.5 to find <span>the average number of cars that pass through the intersection at 6:30 p.m
</span> $y=-6 x^{2} +51.5x+10.2$
$y=-6 (3.5)^{2} +51.5(3.5)+10.2$
$y=-6(12.25)+180.25+10.2$
$y=-73.5+190.45$
$y=116.95$

We can conclude that the <span>average number of cars that pass through an intersection at 6:30 pm is approximately 117. </span>

8 0

2 years ago

Read 2 more answers

From Statistics and Data Analysis from Elementary to Intermediate by Tamhane and Dunlop, pg 265. A thermostat used in an electri

Leviafan [203]

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:

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