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3 years ago

If the ratio is 50g to 300 ml how many ml would be needed for 83g

Mathematics

1 answer:

Natali [406]3 years ago

4 0

Answer:

498 ml would be needed for 83g.

Step-by-step explanation:

♧We can solve using proportion.

$\frac{50 \: g}{300 \: ml} = \frac{83 \: g}{x} \\ \\ \frac{24900}{50} = \frac{50x}{50} \\ \\ 498 = x$

▪Happy To Help <3

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What's the 5% of 180

Vesna [10]

Your answer is 9 hope this helps

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3 years ago

Read 2 more answers

Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

erastovalidia [21]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Using the normal approximation to the binomial.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

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:

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

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.

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:

175 visitors, so $n = 175$

46.7% through the Beaver Meadows, so $p = 0.467$

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

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

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

This is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

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

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

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

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

6.3% over the Grand Lake park entrance, so $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability is P(X < 12 - 0.5) = P(X < 11.5), which is the pvalue of Z when X = 11.5. So

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

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

22.7% with no recorded point, so $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

This probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

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

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

99.78% probability that more than 55 visitors have no recorded point of entry

3 0

3 years ago

james went out for a long walk . He walked 3 2/5 of the mile. then he sat down to take rest . After that he walked 5/7 of a mile

kotykmax [81]

Answer:

Altogether James walked $4\frac{4}{35}$ miles.

Step-by-step explanation:

The question involves mixed fractions. Mixed fractions are first converted into proper fractions and then any operation can be performed on them.

Given

Distance of first walk = $3\frac{2}{5}$ of a mile

Distance of second walk = 5/7

In order to find how much he walked altogether, we have to find the sum of both fractions.

$Total\ distance = 3\frac{2}{5}+\frac{5}{7}\\= \frac{17}{5}+\frac{5}{7}\\=\frac{17*7}{5*7} + \frac{5*5}{7*5}\\=\frac{119}{35} + \frac{25}{35}\\=\frac{119+25}{35}\\=\frac{144}{35}\\=4\frac{4}{35}$