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cestrela7 [59]
2 years ago
6

A 16\dfrac1216 2 1 ​ 16, start fraction, 1, divided by, 2, end fraction kilometer stretch of road needs repairs. Workers can rep

air 2\dfrac142 4 1 ​ 2, start fraction, 1, divided by, 4, end fraction kilometer of road per week. How many weeks will it take to repair this stretch of road?
Mathematics
1 answer:
Solnce55 [7]2 years ago
7 0

Answer:

It will take  7\frac{1}{3} weeks to complete the repair of the stretch of road.

Step-by-step explanation:

Given:

Length of the roads that need repairs = 16\frac{1}{2}\ km

Length of the road workers can repair in a week = 2\frac{1}{4}\ km

To find the number of weeks will it take to repair this stretch of road.

Solution:

Unit rate of workers to repair  roads =   2\frac{1}{4}\ km per week.

Total length of the roads to repair = 16\frac{1}{2}\ km

The number of weeks it will take to repair  16\frac{1}{2}\ km of road by workers can be given as:

⇒ \frac{16\frac{1}{2}\ km}{2\frac{1}{4}\ km}

<em>In order to divide mixed numbers, we convert them first to fractions.</em>

⇒ \frac{\frac{33}{2}} {\frac{9}{4}}

<em>To divide fractions, we take reciprocal of the divisor and replace division with multiplication:</em>

⇒  \frac{33}{2}\times \frac{4}{9}

⇒ \frac{22}{3}

We reconvert the fraction to mixed number.

⇒ 7\frac{1}{3} weeks

Thus, it will take  7\frac{1}{3} weeks to complete the repair of the stretch of road.

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Answer:

8% or 0.08

Step-by-step explanation:

Probability of missing the first pass = 40% = 0.40

Probability of missing the second pass = 20% = 0.20

We have to find the probability that he misses both the passes. Since the two passes are independent of each other, the probability that he misses two passes will be:

Probability of missing 1st pass x Probability of missing 2nd pass

i.e.

Probability of missing two passes in a row = 0.40 x 0.20 = 0.08 = 8%

Thus, there is 8% probability that he misses two passes in a row.

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Levant will receive about 9 american dollars when he exchanges his pesos.

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Answer:

0% probability that at least 1,500 will agree to respond

Step-by-step explanation:

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

Binomial probability distribution

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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:

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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:

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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.

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:

n = 15000, p = 0.09

So

\mu = E(X) = np = 15000*0.09 = 1350

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This is 1 subtracted by the pvalue of Z when X = 1500-1 = 1499. So

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

Z = \frac{1499 - 1350}{35.05}

Z = 4.25

Z = 4.25 has a pvalue of 1.

1 - 1 = 0

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