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Sveta_85 [38]
2 years ago
9

(Will give 1,000 points!) Look at the two 3s in this number:

Mathematics
1 answer:
dusya [7]2 years ago
4 0

The only true statement is <em>"the orange 3 on the right is 1/10 of the value of the blue 3 on the left"</em>

<em />

This question bothers on place values.

<em />

Given the decimal value 89.033, the place value of the blue 3 on the left is the hundredth place (1/100) while the place value of the orange 3 is in the thousandth place(1/1000).

We need to<u> check which of the statement is correct. </u>

For the first option, the blue 3 on the left is 300 times the value of the orange 3 on the right;

  • 1/100 = 300(1/1000)
  • 1/100 = 3/10

Since 0.01≠0.3, hence the first option is wrong

For the second option, the blue 3 on the left is 100 times the value of the orange 3 on the right;

  • 1/100 = 100(1/1000)
  • 1/100 = 1/10

0.01≠0.1, hence the second option is wrong.

The blue 3 on the left is equivalent to the orange 3 on the right is also wrong since they are positioned in a different place

The orange 3 on the right is 1/10 of the value of the blue 3 on the left, is expressed as;

  • 1/1000 = 1/10(1/100)
  • 0.001 = 0.001

Hence the only true statement is <em>"the orange 3 on the right is 1/10 of the value of the blue 3 on the left"</em>

<em />

Learn more on place values here: brainly.com/question/627658

<em />

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3 years ago
. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

Problems of normally distributed samples can be 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 = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

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

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0
3 years ago
Please help me. thanks.
Rama09 [41]

~~~\text{Area of triangle} = \dfrac 12 \times \text{Base} \times \text{Height}\\\\\implies 10 = \dfrac 12 \times 5 \times w\\\\\implies w =\dfrac{20}5 \\\\\implies w = 4 ~cm

8 0
2 years ago
Read 2 more answers
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