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jok3333 [9.3K]
2 years ago
5

1.09g/mL to lbs/gal Please show how to solve it too

Mathematics
2 answers:
kirill115 [55]2 years ago
6 0

Answer : 1.09 g/ml is equal to the 10.9 lbs/gal

Step-by-step explanation :

Conversion factors

1 gram = 0.0022 lbs

1 ml = 0.00022 gallon

First we have to convert 'gram' into'lbs'.

\frac{1.09g}{1ml}=\frac{1.09g}{1ml}\times \frac{0.0022\text{ lbs}}{1g}=\frac{1.09\times 0.0022\text{ lbs}}{1ml}=\frac{0.002398\text{ lbs}}{1ml}

Now we have to convert 'ml' into 'gallon'.

\frac{0.002398\text{ lbs}}{1ml}=\frac{0.002398\text{ lbs}}{1ml}\times \frac{1ml}{0.00022\text{ gallon}}=\frac{10.9\text{ lbs}}{1\text{gallon}}=10.9\text{ lbs}/gallon

Thus, 1.09 g/ml is equal to the 10.9 lbs/gal


GaryK [48]2 years ago
4 0

For this case we must take into account the following conversions:

1 pound = 450 grams

1 mL = 0.00026 gallons

We apply the transformation of units to the following expression:

1.09\frac{g}{mL}

We have then:

(1.09\frac{g}{mL})(\frac{1}{450}\frac{lbs}{g}) = 0.0024\frac{lbs}{mL}

(0.0024\frac{lbs}{mL})(\frac{1}{0.00026}\frac{mL}{gal}) = 9.23\frac{lbs}{gal}

Answer:

Applying both transformations to the given expression we have:

9.23\frac{lbs}{gal}

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Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal to solve this question.

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The expected value of the binomial distribution is:

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The standard deviation of the binomial distribution is:

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Problems of normally distributed samples 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:

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When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.2, n = 200. So

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In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

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