What is the second partial product of 564x307

victus00 [196]

I think answer should be d. Please give me brainlest let me know if it’s correct or not okay thanks bye bye

6 0

3 years ago

The student setup 30 games booths;the teachers also setup some. If 5/6 of the game booths were set up by the students, how many

Ket [755]

25 were set up by the students

6 0

3 years ago

The mean life of a television set is 119 months with a standard deviation of 14 months. If a sample of 74 televisions is randoml

irina [24]

Answer:

50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

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 $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.63$

If a sample of 74 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 1.1 months

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

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

By the Central Limit Theorem

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