An agriculture researcher wants to test the effect of nine soil additives by adding four at a time to different rows of the

Please help me out here!

allsm [11]

9514 1404 393

Answer:

1/4

Step-by-step explanation:

Put the value where the variable is and do the arithmetic.

3z² = 3(√3/6)² = 3(3/6²) = 9/36 = 1/4

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The whole point of √3 is that the square of it is 3. The square of a fraction is the fraction multiplied by itself. The usual rules of multiplying fractions apply.

7 0

2 years ago

Name the property illustrated by each statement:

aniked [119]

both are correct.

....

8 0

3 years ago

Find the area of the figure. The area is square inches.<br><br>please help!

Nostrana [21]

Answer:

154 inchs

Step-by-step explanation:

22 * 14 divided by 2

7 0

3 years ago

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.8. (Round your ans

Alenkinab [10]

Answer:

a) 0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

b) 0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .