Learning Task 1: Give the value and the place value of the underlined

Mathematics

1 answer:

galina1969 [7]3 years ago

5 0

Answer:

Value place value

1. 0.016. hundredths

2. 0009. ten thousandths

3. 00011. ten thousandths

4. 0.00093. ten thousandths

5.0004. ten thousandths

Step-by-step explanation:

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Question 1<br> Determine the midpoint of the segment with endpoint (-9,3) and (7,-8).<br> Question 2

mario62 [17]

Answer:

(-1, -2.5)

Step-by-step explanation:

the midpoint of a segment is $(\frac{x1+x2}{2} ,\frac{y1+y2}{2}$ $)$ , given points (x1, y1) and (x2, y2)

$midpoint=(\frac{-9+7}{2}, \frac{3-8}{2} )$ $= (\frac{-2}{2}, \frac{-5}{2} ) = (-1, -2.5)$

6 0

3 years ago

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