A box turtle marcos found is 0.115 of a meter long. what is this length in expanded form?

Mathematics

2 answers:

Contact [7]3 years ago

8 0

Answer:0.1+.01+0.005

Step-by-step explanation:

egoroff_w [7]3 years ago

5 0

0.1+0.01+0.005 hope this helps

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Find the product of 51 and -2.5. Use the distributive property to rewrite and solve. The first step is done for you. 51(-2.0) +

I am Lyosha [343]

Answer:

-12.75

Step-by-step explanation:

Find the product of 51 and -2.5. Use the distributive property to rewrite and solve.

The next step using Distributive property is to expand the brackets

5.1 × 2.5

Step 1: Distributive property

51(-2.0) + 51(-0.5)

Step 2: Expand the brackets

-10.2 + -2.55

Step 3

- 12.75

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2 years ago

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What is the gcf of 16 and 25

Salsk061 [2.6K]

The GCF is 1. There is no other number that those numbers have in common.

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3 years ago

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A researcher reports survey results by stating that the standard error of the mean is 25 the population standard deviation is 40

bezimeni [28]

Answer:

a) A sample of 256 was used in this survey.

b) 45.14% probability that the point estimate was within ±15 of the population mean

Step-by-step explanation:

This question is solved using the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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 $s = \frac{\sigma}{\sqrt{n}}$ .