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gayaneshka [121]
2 years ago
10

Please Give Information Thx ♡​

Mathematics
1 answer:
Arada [10]2 years ago
6 0
She should choose 2

Explanation
A prime number can only be multiplied by 1 and itself
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Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.
frutty [35]
<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

2^n\leq 2^{n+1}-2^{n-1}-1

where n is a positive integer.

  • Let us take n=1

then we have:

2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2

Hence, the result is true for n=1.

  • Let us assume that the result is true for n=k

i.e.

2^k\leq 2^{k+1}-2^{k-1}-1

  • Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u>  2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Let us take n=k+1

Hence, we have:

2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)

( Since, the result was true for n=k )

Hence, we have:

2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2

Also, we know that:

-2

(

Since, for n=k+1 being a positive integer we have:

2^{(k+1)+1}-2^{(k+1)-1}>0  )

Hence, we have finally,

2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

2^n\leq 2^{n+1}-2^{n-1}-1  where n is a positive integer.

6 0
3 years ago
- Keisha has four letter cards that spell out the word MATH. Keisha picks four cards, one at a
zubka84 [21]

Answer:

1/24

Step-by-step explanation:

1/4 x 1/3 x 1/2 x 1/1

1/24

5 0
2 years ago
Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut
defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125:

This means that n = 125, s = \frac{7320}{\sqrt{125}} = 654.72

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{39000 - 39725}{654.72}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

4 0
2 years ago
If g(x)= 2x-1, and h(x)= square root of x, find (goh)(9)
loris [4]
(goh) = 2(x^1/2)-1
(goh)(9) = 2(9^1/2)-1
= 2(3)-1
= 6-1
= 5
(goh)(9) = 5
5 0
3 years ago
Read 2 more answers
if Ingles 2 and 14 are congruent then line T and then you are parallel what is the theorem to this question ​
Nostrana [21]

Answer:

lol na im good

Step-by-step explanation:

7 0
3 years ago
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