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ikadub [295]
3 years ago
12

The dimensions of the tank are 50 cm by 32 cm by 20 cm. The tank is full of water and sand. The ratio of the volume of water to

the volume of sand is 5 : 1 Work out the number of litres of water in the tank.
Mathematics
1 answer:
Sauron [17]3 years ago
4 0

Given:

Dimensions of a tank are 50 cm by 32 cm by 20 cm

The tank is full of water and sand

Ratio of volume of water to the volume of sand is 5:1

To find:

Volume of water in the tank in liters

Solution:

Without loss of generality, let,

Length of tank = 50 cm

Breadth of tank = 32 cm

Height of tank = 20 cm

Then, we have,

Total Volume of the tank = Length * Breadth * Height

\Rightarrow Total Volume = 50*32*20 cc

\Rightarrow Total Volume = 32000 cc

It is given that,

Volume of water : Volume of sand = 5:1

\Rightarrow (Volume of water)/(Volume of sand) = 5/1

Cross multiplying, we have,

Volume of sand = (Volume of water)/5

It is also given that the tank is full of water and sand. This implies that,

Volume of water + Volume of sand = Total Volume

Using the value, we obtained, we get,

Volume of water + (Volume of water)/5 = Total Volume

\Rightarrow (6 * Volume of water)/5 = Total Volume

Plugging in the value of total volume, we get,

\Rightarrow (6 * Volume of water)/5 = 32000 cc

\Rightarrow Volume of water = \frac{32000*5}{6} cc

\Rightarrow Volume of water \approx 26,666.6667 cc

We know that,

1 cc = 0.001 liters

\Rightarrow 26,666.6667 cc = (26666.6667 * 0.001) liters

\Rightarrow 26,666.6667 cc \approx 26.667 liters

Thus, Volume of water in the tank is approximately 26.667 liters

Final Answer:

Volume of water in the tank (in liters) is approximately 26.667 liters

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Students who have completed a speed reading course have reading speeds that are normally distributed with a mean of 950 words pe
VikaD [51]

Answer:

36.04% probability that at most two of them would read at less than 850 words per minute

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the binomial probability distribution.

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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:

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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.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Percentage of students who read less than 850 words per minute.

Pvalue of Z when X = 850. The mean is \mu = 950 and the standard deviation is \sigma = 200

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

Z = \frac{850 - 950}{200}

Z = -0.5

Z = -0.5 has a pvalue of 0.3085.

30.85 of students read less than 850 words per minute.

If 10 students are selected at random, what is the probability that at most two of them would read at less than 850 words per minute

This is P(X \leq 2) when n = 10, p = 0.3085. So

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{10,0}.(0.3085)^{0}.(0.6915)^{10} = 0.0250

P(X = 1) = C_{10,1}.(0.3085)^{1}.(0.6915)^{9} = 0.1115

P(X = 2) = C_{10,2}.(0.3085)^{2}.(0.6915)^{8} = 0.2239

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0250 + 0.1115 + 0.2239 = 0.3604

36.04% probability that at most two of them would read at less than 850 words per minute

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