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2 years ago

On Monday 4 bricklayers took 3 hours to lay a total of 4200 bricks.

Mathematics

1 answer:

lisabon 2012 [21]2 years ago

7 0

<h2>Hello!</h2>

The answer is:

It will take 4.5 hours for two bricklayers to lay a total of 3150 bricks.

From the statement, we know that on Monday, bricklayers took 3 hours to lay a total of 4200 bricks.

So, we have that the rate of the four bricklayers is given by the following equation:

$Production=Rate*Time\\\\Rate=\frac{Production}{time}$

$Rate=\frac{4200bricks}{3hours}=\frac{1400bricks}{hour}$

Therefore, we have that the rate of the four bricklayers is 1400 bricks per hour, meaning that each bricklayer can lay 350 bricks per hour.

Now, to calculate how many hours it will take to the 2 bricklayers to lay a total of 3150, we need to divide the calculated rate (for bricklayers) by two, since there are only 2 bricklayers working on Tuesday.

So, calculating the time, we have:

$Production=\frac{Rate}{2}*Time$

$3150bricks=\frac{\frac{1400bricks}{hour}}{2}*Time$

$3150bricks*2*\frac{1hour}{1400bricks}=Time$

$6300bricks*\frac{1hour}{1400bricks}=Time$

$Time=4.5hours$

Hence, we have that it will take 4.5 hours for two bricklayers to lay a total of 3150 bricks.

Have a nice day!

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3 years ago

Read 2 more answers

A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu

telo118 [61]

Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

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:

$n = 110, p = 0.2$

$\mu = E(X) = np = 110*0.2 = 22$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952$

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

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

$Z = \frac{25 - 22}{4.1952}$

$Z = 0.715$

$Z = 0.715$ has a pvalue of 0.7626.

X = 17

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

$Z = \frac{17 - 22}{4.1952}$

$Z = -1.19$

$Z = -1.19$ has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

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3 years ago

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Sindrei [870]

Answer:

22.29% probability that both of them scored above a 1520

Step-by-step explanation:

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In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

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The first step to solve the question is find the probability that a student has of scoring above 1520, which is 1 subtracted by the pvalue of Z when X = 1520.

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

$Z = \frac{1520 - 1497}{322}$