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3 years ago

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At a brick laying contest, each layer of bricks is expected to be of an inch tall. Three contestants recorded the fraction of la

yers of brick they placed and the time it took them to place the bricks in the table below.

1 answer:

Ivahew [28]3 years ago

5 0

Answer:

First, determine the number of inches each contestant made by multiplying the fraction of layers by the height of each layer.

Next, determine what fraction of an hour each contestant took to make their fraction of layers by writing each number of minutes as a fraction of 60 minutes.

Last, determine each contestant’s unit rate in inches per hour.

So, Zeke's unit rate was of an inch per hour.

Melanie's unit rate was of an inch per hour.

To find out how much greater Zeke's unit rate was than Xavier's unit rate, subtract Xavier's unit rate from Zeke's unit rate.

Therefore, Zeke's unit rate was of an inch per hour greater than Xavier's unit rate.

Step-by-step explanation:

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What is the distance between 3, 5 and -4, 1

USPshnik [31]

Answer:

Distance = 8.06

Step-by-step explanation:

We have to use the distance formula $d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1})^{2} }$ .

$(x_{1},y_{1}) =$ coordinates of the first point

$(x_{2},y_{2} ) =$ coordinates of the second point

To solve, plug the appropriate values into the formula.

$d=\sqrt{(-4-3)^{2} +(1-5)^{2} } = \sqrt{65} = 8.062257748$

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3 years ago

4n + 5n= -9? What is the answer?

timurjin [86]

add 4n and 5n, which is 9. 9n=-9, and if u divide both sides by nine, n=-1

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3 years ago

Read 2 more answers

(5, 3) and (6, r) m = -1

Rzqust [24]

Answer:

r = 2

Step-by-step explanation:

(2-3)/(6-5)

= -1/1

= -1

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3 years ago

Time spent using e-mail per session is normally distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that

liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

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:

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

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.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 11, \sigma = 3$

a. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 25, s = \frac{3}{\sqrt{25}} = 0.6$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.6}$

$Z = 0.33$

$Z = 0.33$ has a pvalue of 0.6293.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.6}$

$Z = -0.33$

$Z = -0.33$ has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 sessions, what is the probability that the sample mean is between 10.5 and 11 minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

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

$Z = \frac{11 - 11}{0.6}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 10.5

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

$Z = \frac{10.5 - 11}{0.6}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 sessions, what is the probability that the sample mean is between 10.8 and 11.2 minutes?

Here we have that $n = 100, s = \frac{3}{\sqrt{100}} = 0.3$

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

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

By the Central Limit Theorem

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

$Z = \frac{11.2 - 11}{0.3}$

$Z = 0.67$

$Z = 0.67$ has a pvalue of 0.7486.

X = 10.8

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

$Z = \frac{10.8 - 11}{0.3}$

$Z = -0.67$

$Z = -0.67$ has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0

3 years ago

Whats the volume of the prism please show work‼️

attashe74 [19]

Answer:

Step-by-step explanation:

Volume of a prism:

Area of cross section x height

4 0

3 years ago