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lilavasa [31]
3 years ago
10

it takes one fourth of a gallon of paint covers one fifth of a car how much paint covers the whole car​

Mathematics
1 answer:
Komok [63]3 years ago
5 0

Answer:

The volume of paint that covers the whole car is 1\frac{1}{4} gallons or one and a quarter gallons.

Step-by-step explanation:

Let the volume of paint required to cover whole car be 'x'.

Given:

Volume of paint to cover one-fifth of a car = one-fourth of a gallon.

We use unitary method to determine the volume of paint required.

∵ One-fifth of a car requires paint = \frac{1}{4}\ gal

∴ Whole volume of car requires paint = \frac{\frac{1}{4}}{\frac{1}{5}}

=\frac{1}{4}\times \frac{5}{1}\\=\frac{5}{4}\\=1\frac{1}{4}\ gal

Therefore, the volume of paint that covers the whole car is 1\frac{1}{4} gallons or one and a quarter gallons.

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I don’t get exactly how you’re supposed to do this......
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Answer:

9x

Step-by-step explanation:

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Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,
grandymaker [24]

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4 0
2 years ago
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A manufacturing facility produces rolls of tape. Among all of the rolls of tape produced, the mean length of the tape on a roll
Anna [14]

Answer:

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And using the normal atandard table or excel we got:

P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the length of the tape of a population, and for this case we know the following parameters

Where \mu=651.25 and \sigma=0.73

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And using the normal atandard table or excel we got:

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8 0
3 years ago
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