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Dmitry_Shevchenko [17]
3 years ago
12

Researchers suspect that the average number of units earned per semester by college students is rising. A researcher at Calendul

a College wishes to estimate the number of units earned by students during the spring semester at Calendula. To do so, he randomly selects 100 student transcripts and records the number of units each student earned in the spring term. He found that the average number of semester units completed was 12.96 units per student. Identify the population of interest to the researcher.
Mathematics
1 answer:
aleksandrvk [35]3 years ago
7 0

Answer:

All Calendula College students enrolled in the spring.

Step-by-step explanation:

A researcher at Calendula College wishes to estimate the number of units earned by students during the spring semester at Calendula.

To do so, he randomly selects 100 student transcripts from among all Calendula College students enrolled in the spring and records the number of units each student earned in the spring term.

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Given: (a) 2(x+3)=2x+6 and (b) 4(y-3)=4y-12.

To find: The expanded form of the given expressions.

(c) 4(m+n)=4m+4n       (using a(b+c)=ab+ac)

(d) 3(5-q)=15-3q           (using a(b-c)=ab-ac)

(e) 5(2c+1)=10c+5          (using a(b+c)=ab+ac)

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(g) 7(4b-1)=28b-7          (using a(b-c)=ab-ac)

(h) 3(2x+y-5)=3((2x+y)-5)

⇒3(2x+y-5)=3(2x+y)-15         (using a(b-c)=ab-ac)

⇒3(2x+y-5)=6x+3y-15            (using a(b+c)=ab+ac)

(i) 2(6a-4b+3)=2((6a-4b)+3)

⇒2(6a-4b+3)=2(6a-4b)+6         (using a(b+c)=ab+ac)

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⇒ 6(m+n+p)=6(m+n)+6p           (using a(b+c)=ab+ac)

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(l) g(g-3)=g^2-3g           (using a(b-c)=ab-ac)

(m) n(4-n)=4n-n^2        (using a(b-c)=ab-ac)

(n) a(b+c)=ab+ac           (using a(b+c)=ab+ac)

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Step-by-step explanation:

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