IF NOT CORRECT IM REPORTIN YOU!!!!!!!!! What is step 2 in the problem-solving process?

Mathematics

2 answers:

exis [7]2 years ago

6 0

<span>The correct answer is B. Write down the facts and figures.
</span>

Vaselesa [24]2 years ago

3 0

Step 2 would be B. Write down the facts and figures.

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Yeah so can u answer this

Rzqust [24]

Answer:

give me the points its c c c c c c cc c cc c c. c cc cccc c cc c cc

4 0

1 year ago

Read 2 more answers

Help me please i need help read carefully and answer right plz and show proof or evidence

iragen [17]

Answer:

its b because the answer for y needs to change in order for the x column to be true

6 0

2 years ago

Help please anyone up

saul85 [17]

Since area is length times width, it would be the factors of the equation for the area.
X2+11x+28 factors into...
Length is x+7 so...
Width is x+4

7 0

2 years ago

Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1

svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

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