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dangina [55]
3 years ago
9

The ages of some lecturers are 42 54 50 54 50 42 46 46 48 and 48find the mean age​

Mathematics
2 answers:
Virty [35]3 years ago
3 0

Answer:

48

Step-by-step explanation:

add the values of the ages of the lecturers which should give you 434 then divide by the number of lecturers in question

=480/10

=48

PIT_PIT [208]3 years ago
3 0

Answer:

48.

Step-by-step explanation:

There are a total of 10 lecturers in the list

You find the mean by dividing the sum of all the ages by 10.

The mean age = 480/10 = 48.

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Step-by-step explanation:

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What do I do I’m lost I need help?
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27

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the sequence is 4 added to each number. 4 added to 23 is 27

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Nellie took a total of 27 quizzes over the course of 3 weeks. After attending 4 weeks of school this quarter, how many quizzes w
Dennis_Churaev [7]

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The answer is 36

Step-by-step explanation:

If you do 27÷3 you get 9. So if it's 4 weeks you just add 9 which is 36

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3 years ago
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Which of the following graphs shows the preimage P(x)=x^2 and the image I(x)=P(1/3x)?
Ksju [112]

Answer:

The picture with the widest graph in red

Step-by-step explanation:

The graph P(x) is the parent graph for all quadratic functions. It has a vertex of (0,0) and has the following points:

x      f(x)

-2      4

-1       1

0       0

1         1

2       4

The image of l(x) = P(1/3x) changes the points of the function to

x      f(x)

-2/3      4/9

-1/3        1/9

0       0

1/3        1/9

2/3       4/9

This makes the graph much wider. The graph with the widest red graph is the graph.

7 0
3 years ago
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The common ratio of a geometric series is 3 and the sum of the first 8 terms is 3280.
Pavlova-9 [17]

Answer:

The first term of the geometric series is 1

Step-by-step explanation:

In this question, we are tasked with calculating the first term of a geometric series, given the common ratio, and the sum of the first 8 terms.

Mathematically, the sum of terms in a geometric series can be calculated as;

S = a(r^n-1)/( r-1)

where a is the first term that we are looking for

r is the common ratio which is 3 according to the question

n is the number of terms which is 8

S is the sum of the number of terms which is 3280 according to the question

Plugging these values, we have

3280 = a(3^8 -1)/(3-1)

3280 = a( 6561-1)/2

3280 = a(6560)/2

3280 = 3280a

a = 3280/3280

a = 1

6 0
2 years ago
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