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Karo-lina-s [1.5K]
3 years ago
7

Solve 4e^1+3x-9e^5-2x=0

Mathematics
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

its 24x trust me I did this

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6^2-49•1/7<br><br><br> PLEASE HELP <br><br><br> 6^2-49 ( times) 1/7
Masja [62]

The answer is 29 I put it in my scientific calculator

8 0
3 years ago
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A tablet PC contains 3217 music files. The distribution of file size is highly skewed with many small files. Suppose the true me
m_a_m_a [10]

Answer:

Let X the random variable who represents the file sizeof music. We know the following info:

\mu =2.3,\sigma =3.25

We select a sample of n=50 nails. That represent the sample size.  

Since the sample size is large enough n >30, we can use the central limit theorem. From this theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we can approximate the distribution of the sample mean as a normal distribution and no matter if the distribution for X is right skewed or no.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable who represents the file sizeof music. We know the following info:

\mu =2.3,\sigma =3.25

We select a sample of n=50 nails. That represent the sample size.  

Since the sample size is large enough n >30, we can use the central limit theorem. From this theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we can approximate the distribution of the sample mean as a normal distribution and no matter if the distribution for X is right skewed or no.

8 0
3 years ago
The costs the company $0.02 per cubic inch of oatmeal to fill a container. The company does not want the new container to cost m
In-s [12.5K]

Answer:

It is container A

Step-by-step explanation:

just took the quiz

5 0
3 years ago
Read 2 more answers
Three tenths times twelve
Sholpan [36]

Answer:

3.6

Step-by-step explanation:

0.3x12=3.6

Because you know that 12x3=36, so 12x0.3=3.6!

Good luck!

7 0
3 years ago
According to a marketing research study, American teenagers watched 14.8 hours of social media posts per month last year, on ave
ki77a [65]

Answer:

The value of test statistics is 1.06.

Step-by-step explanation:

We are given that according to a marketing research study, American teenagers watched 14.8 hours of social media posts per month last year, on average. A random sample of 11 American teenagers was surveyed and the mean amount of time per month each teenager watched social media posts was 15.6. This data has a sample standard deviation of 2.5.

We have to test if the mean amount of time American teenagers watch social media posts per month is greater than the mean amount of time last year or not.

Let, NULL HYPOTHESIS, H_0 : \mu = 14.8 hours  {means that the mean amount of time American teenagers watch social media posts per month is same as the mean amount of time last year}

ALTERNATE HYPOTHESIS, H_1 : \mu > 14.8 hours  {means that the mean amount of time American teenagers watch social media posts per month is greater than the mean amount of time last year}

The test statistics that will be used here is One-sample t-test;

             T.S. = \frac{\bar X - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, \bar X = sample mean amount of time per month each teenager watched social media posts = 15.6 hours

             s = sample standard deviation = 2.5 hours

             n = sample of teenagers = 11

So, <u>test statistics</u> =  \frac{15.6 - 14.8}{\frac{2.5}{\sqrt{11} } } ~ t_1_0

                            = 1.06

Hence, the value of test statistics is 1.06.

5 0
3 years ago
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