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uysha [10]
3 years ago
14

42 students are going on a field trip. The students are divided into groups of four Students. How many groups of four will there

be. If the remaining studentsform a smaller group, and one chaparone is assigned to every group,how many total chaparones are needed
Mathematics
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

Step-by-step explanation:

if we have 42 students which are divided in group of 4 we will have

42/4=10 groups and 2 students left

if we would have a group out of 3 students, we will have 14 groups and 14 chaperones

a group will have 3 students and one chaperone

42 students and 14 chaperones

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mixas84 [53]

Answer:

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Step-by-step explanation:

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EastWind [94]

The value of n from the given expression is 5.5

<h3>Solution to linear equation and fractions</h3>

Fractions are expression written as a ratio of two integers. The linear equation on the other hand has a leading degree of 1.

Given the equation below;

n/11 = 3.5/7

Simplify to have

n/11 = 1/2

Cross multiply the given result to have:

11 = 2n

Swap

2n = 11

Divide both sides by 2 to have;

2n/2 = 11/2

n = 5.5

Hence the value of n from the given expression is 5.5

Learn more on linear equation and fraction here: brainly.com/question/11955314

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8 0
10 months ago
An experiment to investigate the survival time in hours of an electronic component consists of placing the parts in a test cell
mixer [17]

Answer:

a) The sample mean is of 49 and the sample standard deviation is of 11.7.

b) The range of the true mean at 90% confidence level is of 9.62 hours.

c) The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.

Step-by-step explanation:

Question a:

Sample mean:

\overline{x} = \frac{34+40+46+49+61+64}{6} = 49

Sample standard deviation:

s = sqrt{\frac{(34-49)^2+(40-49)^2+(46-49)^2+(49-49)^2+(61-49)^2+(64-49)^2}{5}} = 11.7

The sample mean is of 49 and the sample standard deviation is of 11.7.

b)Determine the range of the true mean at 90% confidence level.

We have to find the margin of error of the confidence interval. Since we have the standard deviation for the sample, the t-distribution is used.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 6 - 1 = 5

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 5 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.9}{2} = 0.95. So we have T = 2.0.150

The margin of error is:

M = T\frac{s}{\sqrt{n}}

In which s is the standard deviation of the sample and n is the size of the sample. So

M = 2.0150\frac{11.7}{\sqrt{6}} = 9.62

The range of the true mean at 90% confidence level is of 9.62 hours.

(c)If a seventh sample is tested, what is the prediction interval (90% confidence level) of its failure time.

This is the confidence interval, so:

The lower end of the interval is the sample mean subtracted by M. So it is 49 - 9.62 = 39.38 hours.

The upper end of the interval is the sample mean added to M. So it is 49 + 9.62 = 58.62 hours.

The prediction interval, at a 90% confidence level, of it's failure time is between 39.38 hours and 58.62 hours.

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Find the sum of 0.333......and 0.444....​
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Answer:

0.777......

Step-by-step explanation:

0.333...... + 0.444....​..

= 0.777......

Hope it helps :)

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