8 21/40 as a decimal

A three-dimensional array can be thought of as ______ of two-dimensional arrays.

Iteru [2.4K]

A page of two-dimensional arrays can be thought of as a three-dimensional array. Since 2-dimensional arrays are commonly expressed in tables or matrixes, therefore, if we put these tables or matrices in a page, the collection of matrices in a single page would now be structured into a 3D array.

6 0

3 years ago

Estimate -9 1/6 + 2 1/3 + (-1 7/8) I have more questions so if you want more just simply ask.

BigorU [14]

-8.70833333333 about but you can round to the nearest tenth (-8.7) or nearest whole number (-9)

4 0

3 years ago

Translate the word phrase into a math expression.

viktelen [127]

12 meters longer than his throw should be t + 12

t + 12 is your answer

hope this helps

4 0

3 years ago

Read 2 more answers

Nour and Rana are shopping for a Christmas tree. They are deciding between

irga5000 [103]

Answer:

Sample Space:

1. (real, white)

2. (real, silver)

3. (real, gold)

4. (real, purple)

5. (fake, white)

6. (fake, silver)

7. (fake, gold)

8. (fake, purple)

Step-by-step explanation:

Nour and Rana each created a display to represent the sample space of randomly picking a type of tree and a color for the ornaments.

Whose display correctly represents the sample space?

We are not given their sample space but still you can compare the sample space you have with this one to find out whether it is correct or not.

There are 2 different types of Christmas trees

1. Real

2. Fake

There are 4 different types of colors for the ornaments

1. white

2. silver

3. gold

4. purple

The sample space is the set of all the possible outcomes of an experiment.

Then all of the possible outcomes are:

1. (real, white)

2. (real, silver)

3. (real, gold)

4. (real, purple)

5. (fake, white)

6. (fake, silver)

7. (fake, gold)

8. (fake, purple)

Therefore, there are total 8 possible outcomes.

Note: If your sample space matches with this one then it is correct otherwise it is not correct.

4 0

3 years ago

A study of long-distance phone calls made from General Electric's corporate headquarters in Fairfield, Connecticut, revealed the

Jet001 [13]

Answer:

a) 0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

b) 0.0668 = 6.68% of the calls last more than 4.2 minutes

c) 0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

d) 0.9330 = 93.30% of the calls last between 3 and 5 minutes

e) They last at least 4.3 minutes

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 3.6, \sigma = 0.4$

(a) What fraction of the calls last between 3.6 and 4.2 minutes?

This is the pvalue of Z when X = 4.2 subtracted by the pvalue of Z when X = 3.6.

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

X = 3.6

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

$Z = \frac{3.6 - 3.6}{0.4}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.9332 - 0.5 = 0.4332

0.4332 = 43.32% of the calls last between 3.6 and 4.2 minutes

(b) What fraction of the calls last more than 4.2 minutes?

This is 1 subtracted by the pvalue of Z when X = 4.2. So

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% of the calls last more than 4.2 minutes

(c) What fraction of the calls last between 4.2 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 4.2. So

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 4.2

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

$Z = \frac{4.2 - 3.6}{0.4}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

0.9998 - 0.9332 = 0.0666

0.0666 = 6.66% of the calls last between 4.2 and 5 minutes

(d) What fraction of the calls last between 3 and 5 minutes?

This is the pvalue of Z when X = 5 subtracted by the pvalue of Z when X = 3.

X = 5

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

$Z = \frac{5 - 3.6}{0.4}$

$Z = 3.5$

$Z = 3.5$ has a pvalue of 0.9998

X = 3

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

$Z = \frac{3 - 3.6}{0.4}$

$Z = -1.5$

$Z = -1.5$ has a pvalue of 0.0668

0.9998 - 0.0668 = 0.9330

0.9330 = 93.30% of the calls last between 3 and 5 minutes

(e) As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4% of the calls. What is this time?

At least X minutes

X is the 100-4 = 96th percentile, which is found when Z has a pvalue of 0.96. So X when Z = 1.75.

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

$1.75 = \frac{X - 3.6}{0.4}$

$X - 3.6 = 0.4*1.75$

$X = 4.3$

They last at least 4.3 minutes

7 0

2 years ago