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4 years ago

Michael is using a number line to evaluate the expression –8 – 3.

Mathematics

2 answers:

djverab [1.8K]4 years ago

5 0

Answer:

Micheal could rewrite the equation as negative 8 + negative 3 and move 3 spaces to the left.

Step-by-step explanation:

-8 - 3 = -8 + (-3)

ElenaW [278]4 years ago

4 0

Answer:

Here, mark the other dude brainliest.

C. Michael could rewrite the expression as negative 8 + 3 and move 3 spaces to the left.

Answer: -8 - 3 = -8 + (-3)

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3 years ago

Use greatest common factor and the distributive property to write equivalent expressions in factored form for the following expr

Nitella [24]

Answer:

a. 4(d+3e)

b. 6(3x+5y)

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Step-by-step explanation:

In each case we find the greatest common factor of the numbers. That is the greatest number that goes into both the numbers. Then we factor it out in front and inside parentheses we divide each original term by the greatest common factor:

a. 4d+12e, GCF: 4

$\displaystyle4\left(\frac{4d}{4}+\frac{12e}{4}\right)=4(d+3e)$

b. 18x+30y, GCF: 6

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4 years ago

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Answer:

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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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $s = \frac{\sigma}{\sqrt{n}}$ .