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3 years ago

|4x + 8| + 12= 4 I subtracted 12 by 4 and got -8. Would I still solve the equation or would it be no solution?

Mathematics

1 answer:

blsea [12.9K]3 years ago

7 0

Answer:

no solution

Step-by-step explanation:

|4x + 8| + 12= 4

Subtract 12

|4x + 8| = -8

An absolute value must be non negative

Since ours is negative, there is no solution

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Answer:

Step-by-step explanation:

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3 years ago

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aliya0001 [1]

$\bf Step-by-step~explanation:$

If Ava has 34 candy bars, and each box can hold 5 bars, then we need to find out how many boxes that are filled up.

$\bf Step~1:$

Divide the number of candy bars (34), by the number each box can hold (5)

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$\boxed{\bf Our~final~answer:~6~boxes}$

$\bf Check~our~answer:$

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3 years ago

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3 years ago

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