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2 years ago

Answer to this problem 2³×(15-9)+7

Mathematics

2 answers:

Crazy boy [7]2 years ago

6 0

Answer:

The answer is 55.

Step-by-step explanation:

hope this helps

2³×(15-9)+7

=8(15-9)+7

=120-72+7

=48+7

=55

erma4kov [3.2K]2 years ago

5 0

He’s right the awnser is 55

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2 years ago

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1. 8385 double majors

2. 18252 major codes

3. 8515 codes

4. Yes

Step-by-step explanation:

Given

Major areas = 130

Alphabets = 26

There are 130 major areas in the university.

To get a double majors, means one select any of two of the major areas in the university.

i.e. just 2 alphabets are needed

Number of Double Majors = 130C2

130C2 = 130!/(128!*2!)

130C2 = 130 * 129/2

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Number of major codes available = Number of 2 digit codes + Number of 3 digit codes

Number of 2 digit codes = 2 alphabets

There are 26 ways of selecting the first letter

There are 26 ways of selecting the second letter

So, number of 2 digit codes = 26 * 26 = 676.

Number of 3 digit codes = 3 alphabets

There are 26 ways of selecting the first letter

There are 26 ways of selecting the second letter

There are 26 ways of selecting the third letter

So, number of 3 digit codes = 26 * 26 * 26 = 17576

Thus, number of major codes = 17576 + 676 = 18252

Number of codes to identify a student with either a single major or double major

Number off single major codes = number of major areas = 130

Number of double major codes = 8385

Total = 130 + 8385 = 8515

d. From the solutions above, Yes there are enough codes available to identify all single and double majors at the university

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