If f(1) = 2 and f(n) = –2f(n − 1) then find the value of f(5).

The function $y = x ^ n$ , when x approaches infinity, where does y tend? Of course it tends to infinity as well, therefore it has no horizontal asymptotes (and neither vertical nor oblique)

With respect to the function $y= \frac{1}{x}$ we can verify that if it has asymptote horizontal in y = 0. Since when x approaches infinity the function is closer to the value 0.

For example: 1/2 = 0.5; 1/1000 = 0.001; 1/100000 = 0.00001 and so on. As "x" grows "y" approaches zero

Also, when x approaches 0, the function approaches infinity, in other words, when x tends to 0 y tends to infinity. For example: 1 / 0.5 = 2; 1 / 0.1 = 10; 1 / 0.01 = 100 and so on. This means that the function also has an asymptote at x = 0

6 0

3 years ago

Which statement is true?

Lady_Fox [76]

Answer:

An irrational number may show terminating digits after the decimal

Step-by-step explanation:

Examples of Rational Numbers

Number 9 can be written as 9/1 where 9 and 1 both are integers. 0.5 can be written as ½, 5/10 or 10/20 and in the form of all termination decimals. √81 is a rational number, as it can be simplified to 9 and can be expressed as 9/1. 0.7777777 is recurring decimals and is a rational number

Examples of Irrational Numbers

Similarly, as we have already defined that irrational numbers cannot be expressed in fraction or ratio form, let us understand the concepts with few examples.

5/0 is an irrational number, with the denominator as zero.

π is an irrational number which has value 3.142…and is a never-ending and non-repeating number.

√2 is an irrational number, as it cannot be simplified.

0.212112111…is a rational number as it is non-recurring and non-terminating.

There are a lot more examples apart from above-given examples, which differentiate rational numbers and irrational numbers.

Properties of Rational and Irrational Numbers

Here are some rules based on arithmetic operations such as addition and multiplication performed on the rational number and irrational number.

#Rule 1: The sum of two rational numbers is also rational.

Example: 1/2 + 1/3 = (3+2)/6 = 5/6

#Rule 2: The product of two rational number is rational.

Example: 1/2 x 1/3 = 1/6

#Rule 3: The sum of two irrational numbers is not always irrational.

Example: √2+√2 = 2√2 is irrational

2+2√5+(-2√5) = 2 is rational

#Rule 4: The product of two irrational numbers is not always irrational.

Example: √2 x √3 = √6 (Irrational)

√2 x √2 = √4 = 2 (Rational)

5 0

3 years ago

A crossover trial is a type of experiment used to compare two drugs. Subjects take one drug for a period of time and then switch

zysi [14]

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369$

$df=n-1=7-1=6$

$p_v =2*P(t_{(6)}$

We see that the p value is higher than the ususal significance levels commonly used of 1% or 5% so then we can conclude that we FAIL to reject the null hypothesis, and there is not enough evidence to conclude that we have a different response between the two drugs

Step-by-step explanation:

We have the following info given by the problem

Subject 1 2 3 4 5 6 7

Drug A 6 3 4 5 7 1 4

Drug B 5 1 5 5 5 2 2

x=value for drug A , y = value for drug B

x: 6 3 4 5 7 1 4

y: 5 1 5 5 5 2 2

We want to verify if the mean response differs between the two drugs then the system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x = 0$

Alternative hypothesis: $\mu_y -\mu_x \neq 0$

We can begin calculating the difference $d_i=y_i-x_i$ and we obtain this:

d: -1, -2, 1, 0, -2, 1, -2

Now we can calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=-0.714$

Now we can find the the standard deviation for the differences, and we got:

$s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}}=1.38$

And now we can calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-0.714 -0}{\frac{1.38}{\sqrt{7}}}=-1.369$

Now we can find the degrees of freedom given by:

$df=n-1=7-1=6$

We can calculate the p value, since we have a two tailed test the p value is given by:

$p_v =2*P(t_{(6)}$

We see that the p value is higher than the ususal significance levels commonly used of 1% or 5% so then we can conclude that we FAIL to reject the null hypothesis, and there is not enough evidence to conclude that we have a different response between the two drugs

7 0

3 years ago