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1 year ago

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If a sample of 362 customers is taken from a population of 5530 customers,

1 answer:

inysia [295]1 year ago

6 0

Using concepts of <u>sample and population</u>, it is found that the sample variance is representative of 362 and 5530 customers ages, option D.

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In sampling, the <u>information is taken from a sample</u>, and is used to <u>estimate it for the whole population</u>.

In this problem, we have a sample of 362 and a population of 5530 customers.
The sample variance $s^2$ is calculated from the sample, and used as an estimate for the population variance. Thus, it can be said that it represents both 362 and 5530 customers, and the correct option is D.

A similar problem is given at brainly.com/question/4086221

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Identify the GCF of 6x^2y^2 − 8xy^2 + 10xy

myrzilka [38]

Answer:

The GCF for the numerical part is 2

Step-by-step explanation:

6x^2y^2-8xy^2+10xy

It contains both numbers and variables, there are two steps to find the GCF(HCF).

1). Find the GCF for the numerical part 6, -8,10

2). Find the GCF for the variable part x^2,y^2,x^1,y^2,x^1,y^3

3).Multiply the values together.

Find the common factors for the numerical part:

6,-8,10

Factors of 6

6: 1,2,3,6

Factors of -8

-8: -8,-4,-2,-1,1,2,4,8

Factors of 10

10:1,2,5,10

Common factors of 6,-8, 10 are 1,2

The GCF Numerical=2

The GCF Variable= xy^2

Multiply the GCF of the numerical part 2 and the GCF of the variable part xy^2, and you'll get 2xy^2

6 0

2 years ago

Andrew is about to leave for school. If he walks at a speed of 50 meters per minute, he will arrive 3 minutes after the bell rin

Mariana [72]

Answer:

The answer is: 13 minutes

Step-by-step explanation:

First Let us form equations with the statements in the two scenario

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

Let the time in which the bell rings be 'x'

1. If Andrew walks (50 meters/minute), he arrives 3 minutes after the bell rings. Therefore the time of arrival at this speed = (3 + x) minutes

$3 + x =\frac{distance}{50}\\distance = 50(3+x) - - - - - (1)$

2. If Andrew runs (80 meters/minute), he arrives 3 minutes before the bell rings. Therefore the time taken to travel the distance = (x - 3) minutes

$x - 3 = \frac{distance}{80} \\distance = 80(x-3) - - - - - (2)$

In both cases, the same distance is travelled, therefore, equation (1) = equation (2)

$50(3+x)=80(x-3)$

$150 +50x=80x-240\\$

Next, collecting like terms:

$150 + 240 = 80x - 50x\\390 = 30x\\30x = 390\\$

dividing both sides by 3:

x = 390÷30 = 13

∴ x = 13 minutes

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3 years ago