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tresset_1 [31]
3 years ago
5

A large aquarium contains only two kinds of fish, guppies and swordtails. If 3/4​​ of the number of guppies is equal to 2/3​​ of

the number of swordtails, then what fraction of fish in this aquarium are guppies?
Mathematics
1 answer:
Jlenok [28]3 years ago
6 0

Answer:

\frac{8}{17} of fish in this aquarium are guppies.

Step-by-step explanation:

Let x be the number of guppies and y be the number of swordtails in the aquarium,

According to the question,

\frac{3}{4}\text{ of } x=\frac{2}{3}\text{ of }y

\frac{3x}{4}=\frac{2y}{3}

By cross multiplication,

9x=8y

\implies \frac{x}{y}=\frac{8}{9}

Thus, the ratio of guppies and swordtail fishes is 8 : 9

Let guppies = 8x, swordtail = 9x

Where, x is any number,

Since, the aquarium contains only two kinds of fish, guppies and swordtails,

So, the total fishes = 8x + 9x = 17x

Hence, the fraction of fish in the aquarium are guppies = \frac{\text{Guppies}}{\text{Total fishes}}

=\frac{8x}{17x}

=\frac{8}{17}

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The mean annual cost of an automotive insurance policy is normally distributed with a mean of $1140 and standard deviation of $3
DerKrebs [107]

Using the normal distribution, it is found that the probabilities are given as follows:

a) 0.8871 = 88.71%.

b) 0.0778 = 7.78%.

c) 0.8485 = 84.85%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

The parameters in this problem are given as follows:

\mu = 1140, \sigma = 310, n = 16, s = \frac{310}{\sqrt{16}} = 77.5

Item a:

The probability is the <u>p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1000</u>, hence:

X = 1250:

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

By the Central Limit Theorem

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

Z = \frac{1250 - 1140}{77.5}

Z = 1.42

Z = 1.42 has a p-value of 0.9222.

X = 1000:

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

Z = \frac{1000 - 1140}{77.5}

Z = -1.81

Z = -1.81 has a p-value of 0.0351.

0.9222 - 0.0351 = 0.8871 = 88.71% probability.

Item b:

The probability is <u>one subtracted by the p-value of Z when X = 1250</u>, hence:

1 - 0.9222 = 0.0778 = 7.78%.

Item c:

The probability is the <u>p-value of Z when X = 1220</u>, hence:

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

Z = \frac{1220 - 1140}{77.5}

Z = 1.03

Z = 1.03 has a p-value of 0.8485.

0.8485 = 84.85% probability.

More can be learned about the normal distribution at brainly.com/question/4079902

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Step-by-step explanation:

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