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mafiozo [28]
3 years ago
6

Which equation is the inverse of (x-4)2 - 5 = 6y-12​

Mathematics
1 answer:
qaws [65]3 years ago
3 0

Answer:

hey the answer is b the explanation for that is that i hate baby yoda any one like jake paul because i do i wonder why my parents dont love me help help help

Step-by-step explanation:

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3 years ago
Simplify the following:<br> 3÷4i 15÷2-3i
seropon [69]

Answer:

7i=4.5

Step-by-step explanation:

Hope this helps

7 0
3 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
What is this plxxxxx
Thepotemich [5.8K]

She pays 9.6 euros. Hope this helped, could I possibly get brainliest?

6 0
3 years ago
Read 2 more answers
What is the effect on the graph of the function f(x) = x^2 when f(x) is changed to f(x − 6)
olya-2409 [2.1K]

Answer:

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing f(x) = x^2 and \:f\left(x\right)=\left(x-6\right)^2 is represented as being shifted 6 units to the right as compare to the function f(x) = x^2.

Step-by-step explanation:

When we Add or subtract a positive constant, let say c, to input x, it would be a horizontal shift.  

For example:

Type of change               Effect on y = f(x)

y = f(x - c)                      horizontal shift: c units to right

So

Considering the function

f(x) = x^2

The graph is shown below. The first figure is representing f(x) = x^2.

Now, considering the function

\:f\left(x\right)=\left(x-6\right)^2

According to the rule, as we have discussed above, as a positive constant 6 is added to the input, so there is a horizontal shift, 6 units to the right.

The graph of \:f\left(x-6\right)=\left(x-6\right)^2 is shown below in second figure. It is clear that the graph of  \:f\left(x-6\right)=\left(x-6\right)^2  is shifted 6 units to the right as compare to the function f(x) = x^2.

The comparison graph is also attached in 3rd figure. In the 3rd figure, the graph with vertex (0, 0) is representing f(x) = x^2 and \:f\left(x\right)=\left(x-6\right)^2 is represented as being shifted 6 units to the right as compare to the function f(x) = x^2.

5 0
3 years ago
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