Use a truth table to determine wether the symbolic form of the argument is valid or invalid :

Un agricultor observó que la expresión F(x)=-2x2+16x+25 describe la producción (P), en toneladas, de cacao que cosecha en sus ti

faltersainse [42]

Answer:

348 7gb93c9 good morning to you all right place on my bed right now iam not geting the best way u are gonna be in this situation is it was not Halloween but gug batson vocal vox's John b bucko and my dear loving n love u soo hapy that u can go ahead with it is my beauty grl for the kichen and u will not cm to ur 2AM to go to Gangtok and my 7th anniversary of the 4588th 468inches I will not be the 6th of God Amen n and 93feet

Step-by-step explanation:

galvanise and I will have the best of my kds to 67055965th in the name and u have to write a book about this balance of God and blow it up n and then be online til now I will be a man fan of u my son Anup he shall wbsoenndkd6543 the best man in the land and my 8AM shall il be bsy in the 34th century abksbdkfusnliqnksus in jsbsbdbsmqjnans bus and

6 0

2 years ago

According to a recent study, some experts believe that 15% of all freshwater fish in a particular country have such high level

Lera25 [3.4K]

Answer:

The approximate probability that the market will have a proportion of fish with dangerously high levels of mercury that is more than two standard errors above 0.15 is 0.95.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

$\mu_{\hat p}=0.15$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

As the sample size is large, i.e. <em>n</em> = 150 > 30, the central limit theorem can be used to approximate the sampling distribution of sample proportion by the normal distribution.

Compute the mean and standard deviation as follows:

$\mu_{\hat p}=0.15\\\\\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.15(1-0.15)}{150}}=0.0292$

So, $\hat p\sim N(0.15, 0.0292^{2})$

In statistics, the 68–95–99.7 rule, also recognized as the empirical rule, is a shortcut used to recall that 68%, 95% and 99.7% of the Normal distribution lie within one, two and three standard deviations of the mean, respectively.

Then,

P (µ-σ < X < µ+σ) ≈ 0.68

P (µ-2σ <X < µ+2σ) ≈ 0.95

P (µ-3σ <X < µ+3σ) ≈ 0.997

Then the approximate probability that the market will have a proportion of fish with dangerously high levels of mercury that is more than two standard errors above 0.15 is 0.95.

That is:

$P(\mu_{\hat p}-2\sigma_{\hat p}$

4 0

3 years ago

A distribution of values is normal with a mean of 195.2 and a standard deviation of 68.7. Find P6, which is the score separating

weeeeeb [17]

Answer:

P6 = 88.3715

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 195.2, \sigma = 68.7$