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dexar [7]
3 years ago
9

After raining for 34 of an hour, a rain gauge is 25 filled. If it continues to rain at that rate for 15 more minutes, what fract

ion of the rain gauge will be filled?
To help answer this question, Diego wrote the division equation 34÷25=?. Explain why this equation does not represent the situation.
Write a multiplication equation and a division equation that does represent the situation.
Mathematics
1 answer:
ivann1987 [24]3 years ago
4 0

Answer:

\textbf{Fraction of gauge filled = }\bf\frac{8}{15}

Step-by-step explanation:

This equation does not represent the situation because to find the fraction of the rain gauge we need to : divide the fraction of gauge filled by the raining fraction of an hour

But, Diego wrote the incorrect division equation so, it does not represent the current situation.

Now, to find the required division and multiplication equations :

\frac{3}{4}\text{ fraction of an hour = }\frac{3}{4}\times 60=\text{ 45 minutes}\\\\\text{ Now, the rain continues to rain for 15 minutes more}\\\implies \text{The total raining time = 45 + 15 = 60 minutes}\\\\\text{So, we need to find the fraction of gauge filled in 60 minutes}\\\\\text{Fraction of gauge filled in 45 minute = }\frac{2}{5}\\\\\text{Fraction of gauge filled in 1 minute = }\frac{\frac{2}{5}}{45}=\frac{2}{5\times 45}\\\\\text{Fraction of gauge filled in 60 minutes = }\frac{2}{5\times 45}\times 60=\bf\frac{8}{15}

\text{Division equation = }\frac{\text{fraction of gauge filled}}{\text{raining fraction of an hour}}\\\\\textbf{Division Equation : }\frac{\frac{2}{5}}{\frac{3}{4}}=\frac{2}{5}\times \frac{4}{3}=\frac{8}{15}\\\\\text{Multiplication Equation = Fraction of gauge filled in 1 minute × 60}\\\\\textbf{Multiplicatiuon Equation : }\frac{2}{225}\times 60=\frac{8}{15}

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Consider the probability that no less than 92 out of 159 students will pass their college placement exams. Assume the probabilit
Yanka [14]

Answer:

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

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:

p = 0.55, n = 159. So

\mu = E(X) = 159*0.55 = 87.45

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{159*0.55*0.45} = 6.27

Probability that no less than 92 out of 159 students will pass their college placement exams.

No less than 92 is more than 91, which is 1 subtracted by the pvalue of Z when X = 91. So

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

Z = \frac{91 - 87.45}{6.27}

Z = 0.57

Z = 0.57 has a pvalue of 0.7157

1 - 0.7157 = 0.2843

0.2843 = 28.43% probability that no less than 92 out of 159 students will pass their college placement exams.

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3 years ago
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