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Mathematics

2 answers:

GrogVix [38]3 years ago

6 0

The answer is the 3 but three lines close to the 4 if that makes sense

ra1l [238]3 years ago

3 0

Well the square root of 14 is 3.7 so i would go to the 3 then go to the 7th tic mark????? hope this helps

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In a standard normal distribution, above which z-score do approximately 63% of the date lie? A.) 1.79 B.)-0.33 C.) -1.79 D.)0.33

Dovator [93]

If 63% lie above the z-score, then 37% lie below, or to the left. On the "negative" chart locate the value closest to 0.37. It is 0.3070 that corresponds to the z-score of -0.33.

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3 years ago

The softball team has won 7 of their last 8 games.

Inga [223]

Answer: 88%

Step-by-step explanation:

(7/8)*100=87.5

Which is approximately 88.

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3 years ago

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5. Let f(x) = 3x3/4 and g(x) = -2x314 Find (f + g)(x). Please show work.

pishuonlain [190]

Answer:

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Step-by-step explanation:

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3 years ago

A process is producing a particular part where the thickness of the part is following a normal distribution with a µ = 50 mm and

Hitman42 [59]

Answer:

0.13% probability that this selected sample has an average thickness greater than 53

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 $s = \frac{\sigma}{\sqrt{n}}$ .