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RSB [31]
3 years ago
13

i don't know if this makes sense but I have to make 8X to the 12 power a number in parentheses. I have to fill in the boxes with

a different number​

Mathematics
1 answer:
Scilla [17]3 years ago
8 0

\bf 8x^{12}\qquad \begin{cases} 8=&2\cdot 2\cdot 2\\ &2^3\\ x^{12}=&x^{4\cdot 3}\\ &(x^4)^3 \end{cases}\implies 2^3(x^4)^3\implies \left( 2x^4 \right)^3

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Solve y = x + 6 for X.<br> x= y + 6<br> x= -y + 6<br> x = y - 6<br> x= -y - 6
damaskus [11]

Answer:

x=y-6

Step-by-step explanation:

First, flip the equation.

x+6=y

Finally, add -6 to both sides.

x+6+−6=y+−6

x=y−6

6 0
2 years ago
Read 2 more answers
How do you solve this equation?
Natasha2012 [34]

Answer:

5 \sqrt{2}

= 7.071067812

Step-by-step explanation:

hope this helps

8 0
2 years ago
A tobacco company claims that the amount of nicotene in its cigarettes is a random variable with mean 2.2 and standard deviation
Aleksandr-060686 [28]

Answer:

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

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 normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 2.2, \sigma = 0.3, n = 100, s = \frac{0.3}{\sqrt{100}} = 0.03

What is the approximate probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true?

This is 1 subtracted by the pvalue of Z when X = 3.1. So

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

By the Central Limit Theorem

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

Z = \frac{3.1 - 2.2}{0.03}

Z = 30

Z = 30 has a pvalue of 1.

1 - 1 = 0

0% probability that the sample mean would have been as high or higher than 3.1 if the company’s claims were true.

4 0
3 years ago
You purchase one microsoft july $72 put contract for a premium of $1. 32. what is your maximum possible profit?
goldenfox [79]

The maximum possible profit = $7068

For given question,

One Microsoft July $72 put contract for a premium of $1.32

The payoff arise from put option is max (K - S, 0) - P

Now it would be maximum at S = 0

And, the maximum payoff is

K - 0 - P

= K - P

= 72 - 1.32

= $70.68

We assume that for each and every contract the number of shares is 100

So, the maximum profit gained from this strategy is

= $70.68 × 100 shares

= $7068

The maximum profit that will be gained from this strategy is $7068

Therefore, the maximum possible profit = $7068

Learn more about the profit here:

brainly.com/question/20165321

#SPJ4

4 0
1 year ago
I need help with this 20 points
iren2701 [21]

Answer:

I dont understand that sorry

4 0
3 years ago
Read 2 more answers
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