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Nutka1998 [239]
3 years ago
6

In a mathematics class, half of the students scored 79 on an achievement test. With the exception of a few students

Mathematics
1 answer:
Yakvenalex [24]3 years ago
3 0
In this situation, you have:
The mode is 79 (that’s the most common number).
The median is 78.5 (the middle value or the average of the two middle values, since there is an even number of test scores).
The mean is less than both of those, as it was brought down by the few scores of 49.

The means the mean will be less than median.

Answer: C
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ε = {x: 2 x 30, x is an integer}, M = {even numbers}, P = {prime numbers}, T = {odd numbers} Find: I) MUP ii) M - T iii) P(MT) i
mariarad [96]

Answer:

Step-by-step explanation:

ε = {x: 2≤ x ≤30}

M = { even numbers} = {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}

P = { prime numbers} = {2,3,5,7,11,13,17,19,23,29}

T = {odd numbers} = {3, 5, 7, 9, 11,13,15,17,19,21,23,25,27,29}

1. M ∪ P = {2,3,4,5,6,7,8,10,11,12,13,14,16,17,18,19,20,22,23,24,26,28,29,30}

2. M - T =

n(M) - n(T)

15- 14 = 1

3. P(MT)

(MT) = M ∩ T = 0

P ∪ (M ∩ T ) = {2,3,5,7,11,13,17,19,23,29}

4. P' = not a prime number

T' = not odd number = M

P' ∪(M∩T')

P' ∪ {2,4,6,8,10,12,14,16,18,20,22,24,26,28,30}

= {2,4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30}

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3 years ago
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Therefore it is C
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