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AnnZ [28]
1 year ago
8

If you have 60 million barrels and you have 14% kerosene and 34% gasoline what is the total production of gasoline and kerosene

combined barrels
Mathematics
2 answers:
Drupady [299]1 year ago
6 0
48% or 48/100 only in percents and fractions
In-s [12.5K]1 year ago
4 0

Answer:

Total production of gasoline and kerosene combined 48% that is 28.8 million.

Step-by-step explanation:

Total production = 60 million barrels

Gasoline = 34%

Kerosene = 14%

Total combined = 34% + 14% = 48%

48% of 60 million = 0.48 × 60 = 28.8 million

Total production of gasoline and kerosene combined 48% that is 28.8 million.

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An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t
Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

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 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 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}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

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

By the Central Limit Theorem

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

Z = \frac{246 - 245}{2.61}

Z = 0.38

Z = 0.38 has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

4 0
2 years ago
Just need to check these answers
amid [387]
Great Job! they are all correct.  :)


Good luck in your next tests.
3 0
2 years ago
Simplify 14.2 + 6.5(p + 3q) – 4.05q
anzhelika [568]
Answer is C
14.2+6.5(p+3q)-4.05q
14.2+6.5p+19.5q-4.05q
14.2+6.5p+15.45q→6.5p+15.45q+14.2
so the answer is C
7 0
1 year ago
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