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3 years ago

9

Serena has several package of raisins. each package contains 3 boxes of raisins. which could be the number of boxes of raisins s

erena has? mark all that
apply

a.9
b.18
c.23
d.26
e.32

GIVE U A LOT OF POINTS!!!!!!!!!!!!!!!!

2 answers:

natta225 [31]3 years ago

5 0

It's A and B because you can divide 9 and 18 evenly into 3 without any remainders :D

Ne4ueva [31]3 years ago

4 0

9
18
are the answers. Hope this helps! -J

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3 years ago

Read 2 more answers

El recorrido de una etapa de una vuelta ciclística tiene una longitud de 213 kilometros y un ciclista recorre 2/3 del trayecto e

Maurinko [17]

the question in English

The route of a stage of a cycle tour is 213 km long and a cyclist covers 2/3 of the route in five hours.

a) How many kilometers are left to finish the stage?

b) If he continues with the same average speed, how much time does he have left to finish the stage?

Step 1

we know that

1) Total kilometers of a stage is $213\ km$

2) The cyclist covers $\frac{2}{3}$ of the route

so

the kilometers remaining to finish the stage is equal to

$(1-\frac{2}{3})*213= \frac{1}{3}*213\\ \\=71\ km$

therefore

<u>the answer Part a) is </u>

$71\ km$

Step 2

<u>Find the average speed of the cyclist</u>

we know that

the speed is equal to

$speed=\frac{distance}{time}$

we have

$time=5\ hours\\ distance=\frac{2}{3}*213=142\ km$

substitute

$speed=\frac{142}{5}=28.4\frac{km}{h}$

Step 3

<u>Find the time remaining to finish the stage</u>

we know that

$speed=\frac{distance}{time}$

Solve for the time

$time=\frac{distance}{speed}$

we have

$distance=71\ km\\ speed=28.4\frac{km}{h}$

substitute

$time=\frac{71}{28.4}=2.5\ hours$

therefore

<u>the answer Part b) is</u>

$2.5\ hours$

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3 years ago

To teach computer programming to employees, many firms use on the job training. A human resources administrator wishes to review

tatiyna

Answer:

The z-score for the trainee is of 2.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The mean of the test scores is 72 with a standard deviation of 5.

This means that $\mu = 72, \sigma = 5$

Find the z-score for a trainee, given a score of 82.

This is Z when X = 82. So

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

$Z = \frac{82 - 72}{5}$

$Z = 2$

The z-score for the trainee is of 2.

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2 years ago