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3 years ago

A function is given. Determine the average rate of change of the function between the given values of the variable: f(x)= 4x^2;

x=3, x=3+h

Mathematics

1 answer:

nevsk [136]3 years ago

8 0

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------$

$\bf f(x)= 4x^2 \qquad 
\begin{cases}
x_1=3\\
x_2=3+h
\end{cases}\implies \cfrac{f(3+h)-f(3)}{(3+h)~-~(3)}
\\\\\\
\cfrac{[4(3+h)^2]~~-~~[4(3)^2]}{\underline{3}+h-\underline{3}}\implies \cfrac{4(3^2+6h+h^2)~~-~~4(9)}{h}
\\\\\\
\cfrac{4(9+6h+h^2)~~-~~36}{h}\implies \cfrac{\underline{36}+24h+4h^2~~\underline{-~~36}}{h}
\\\\\\
\cfrac{24h+4h^2}{h}\implies \cfrac{\underline{h}(24+4h)}{\underline{h}}\implies 24+4h$

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A jar contains candies and chocolates. For every 9 candies in the jar, there are 7 chocolates. If there are 336 items in the jar

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3 years ago

Consider the probability that exactly 20 out of 309 people will get the flu this winter. Choose the best description of the area

noname [10]

Answer:

The distribution of proportion of people who will get the flu this winter is N (0.065, 0.014²).

Step-by-step explanation:

Let X = number of people who will get flu this year.

The sample selected is of size, n = 309.

The number of people who will get flu in this sample is, x = 20.

Compute the sample proportion of people who will get flu as follows:

$p=\frac{x}{n}=\frac{20}{309}=0.065$

The random variable X follows a Binomial distribution with parameters n = 309 and p = 0.065.

The sample size is quite large, i.e. n = 309 > 30.

And the probability of success is low.

So the Normal approximation to Binomial can be used to approximate the distribution of sample proportion is:

np ≥ 10
n(1 - p) ≥ 10

Check whether the conditions are fulfilled or not as follows:

$np=309\times 0.065=20.085>10\\n(1-p)=309\times (1-0.065)=288.915>10$

Hence, the conditions are fulfilled.

The sampling distribution of sample proportion is:

$p=N(p, \frac{p(1-p)}{n})$

Compute the mean and variance as follows:

$Mean=p=0.065\\Variance=\frac{p(1-p)}{n}=\frac{0.065(1-0.065)}{309}=0.014^{2}$

Thus, the distribution of proportion of people who will get the flu this winter is N (0.065, 0.014²).

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