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3 years ago

Why mathematical induction is used ?(write a 100 words paragraph)

Mathematics

1 answer:

weeeeeb [17]3 years ago

4 0

Answer:

Mathematical induction is a method of mathematical proof typically used to establish that a given statement is true for all natural numbers ( non-negitive integers). The simplest and most common form of mathematical induction proves that a statement involving a natural number that holds for all values.

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Simplify the following polynomial and write the answer in standard form. (-2x^3+5x^2-4x+8)-(-2x^3+2x-3)

Masja [62]

$(-2x^3+5x^2-4x+8)-(-2x^3+2x-3)\\\\=-2x^3+5x^2-4x+8-(-2x^3)-2x-(-3)\\\\=-2x^3+5x^2-4x+8+2x^3-2x+3\\\\=(-2x^3+2x^3)+5x^2+(-4x-2x)+(8+3)\\\\=5x^2-6x+11$

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3 years ago

Consider the function g(x)=3x+2x-4 (a)(-2) (b)(0) (c)(4) (d)(15)

WINSTONCH [101]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou

Fed [463]

Answer:

$\displaystyle P(A)=\frac{6}{11}$

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

$\Omega=\{WW,WN,NW,NN\}$

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

$A=\{WN,NW\}$

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

$\displaystyle \frac{5}{11}$

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

$\displaystyle \frac{6}{10}$

Thus the option WN has the probability

$\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}$

Now for the second option NW. The initial probability to pick a non-white counter is

$\displaystyle \frac{6}{11}$

The probability to pick a white counter is

$\displaystyle \frac{5}{10}$

Thus the option NW has the probability

$\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}$

The total probability of event A is the sum of both

$\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}$

$\boxed{\displaystyle P(A)=\frac{6}{11}}$

7 0

3 years ago

An automobile manufacturer who wishes to advertise that one of its models achieves 30 mpg (miles per gallon) decides to carry ou

Katyanochek1 [597]

Answer:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$

$p_v =P(t_{(5)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation

The mean and sample deviation can be calculated from the following formulas:

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

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}$

$\bar X=29.35$ represent the sample mean

$s=1.365$ represent the sample standard deviation

$n=6$ sample size

$\mu_o =30$ represent the value that we want to test

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

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 30 or no, the system of hypothesis are :

Null hypothesis: $\mu \geq 30$

Alternative hypothesis: $\mu < 30$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$

P-value

We need to calculate the degrees of freedom first given by:

$df=n-1=6-1=5$

Since is a one-side left tailed test the p value would given by:

$p_v =P(t_{(5)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

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3 years ago

Sam needs to rent a van for a school field trip. Van Company A charges a one time fee of $250 plus $10 for each mile driven. Van

Sonbull [250]

D.y = 10x + 250

y = 12x + 150 is the correct answer to the question

6 0

3 years ago

Why mathematical induction is used ?(write a 100 words paragraph)​

Why mathematical induction is used ?(write a 100 words paragraph)