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Blizzard [7]
3 years ago
15

What is the Absolute value 18/8

Mathematics
1 answer:
vlada-n [284]3 years ago
6 0
18/8
8/8

1/8 = 0.125

1+0.125=1.125
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Multiplying monomials and binomials
lubasha [3.4K]

Answer:

28w^2-476w

Step-by-step explanation:

The general rule we are going to use to multiply this out is the distributive property. Which is:

a(b+c) = ab + ac

<u>Note:</u> x * x = x^2

Now multiplying, we get:

28w(w-17)\\=28w*w-28w*17\\=28w^2-476w

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3 years ago
Shyla had a balance of −$255.25 in her account, and Susan had a balance of −$178.75 in her account. Which statement explains why
Nesterboy [21]

It is given that, Shyla had a balance of −$255.25 in her account, and Susan had a balance of −$178.75 in her account.

And absolute value of -255.25 =255.25 which is greater than absolute value of -178.75 which is 178.75. It means , Shyla's account balance is more than Susan's account balance . THerefore, Shyla owes the bank more than Susan .


5 0
3 years ago
The mean weight of boxes shipped by a company is 12 lbs., with a standard deviation of 4 lbs. Assume weight of boxes follows Nor
777dan777 [17]

Answer:

0.89% probability that a palette of 10 boxes will exceed that limit (150 lbs.)

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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 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, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 12, \sigma = 4, n = 10, s = \frac{4}{\sqrt{10}} = 1.2649

1. What’s the probability that a palette of 10 boxes will exceed that limit (150 lbs.)? a. HINT: Prob (total weight of a sample of 10 boxes >150 lbs.) = Prob (the sample mean weight ( y hat ) > 15 lbs.)

This is 1 subtracted by the pvalue of Z when X = 15. So

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

By the Central Limit Theorem

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

Z = \frac{15 - 12}{1.2649}

Z = 2.37

Z = 2.37 has a pvalue of 0.9911

1 - 0.9911 = 0.0089

0.89% probability that a palette of 10 boxes will exceed that limit (150 lbs.)

6 0
4 years ago
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